Lesson Explainer: Optimization: Applications on Extreme Values Mathematics • Higher Education

In this explainer, we will learn how to apply derivatives to real-world problems to find the maximum and the minimum values of a function under certain conditions.

The basic idea of all optimization problems is the same; we have a particular quantity, such as cost, area, or volume, that we want to minimize or maximize. That quantity is a dependent variable described by a function of the independent variables, such as time, length, or temperature.

Specific real-world examples of optimization problems include a company wanting to minimize production costs or maximize revenue or profit. In manufacturing, it is often desirable to minimize the amount of material used to package a product with a certain fixed volume; boxes of different dimensions can have the same volume but different surface areas. Many optimization problems also have an additional auxiliary condition or constraint, relating the independent variables, that we have to take into account. For example, we may want to maximize the area of a rectangular garden with a particular constraint on how much wire fencing we have for three sides of the rectangle.

In a real-world optimization problem, the first step is to specify the dependent and independent variables; a diagram may also help to visualize the problem. We then determine which quantity we want to minimize or maximize and express that quantity as a function to be optimized, including any additional constraints relating the independent variables. We then determine the critical (or stationary) points of the function using the first derivative test and classify them as (local) maxima or minima by the second derivative test.

Let’s first remind ourselves of the definition of a critical point.

Definition: Critical Point

The point (𝑐,𝑓(𝑐)) is called a critical point of 𝑦=𝑓(π‘₯), provided π‘₯=𝑐 is in the domain of the function 𝑓(π‘₯) and 𝑓(π‘₯) is continuous at π‘₯=𝑐, and either 𝑓′(𝑐)=0𝑓′(𝑐).orisundefined

A critical point can be classified as either a minimum or maximum according to the second derivative test as follows.

Definition: Second Derivative Test

Suppose 𝑓 is a twice differentiable function defined on some interval containing the critical point π‘₯=𝑐. Then

  • if 𝑓′′(𝑐)<0, then 𝑓(π‘₯) has a local maximum at π‘₯=𝑐;
  • if 𝑓′′(𝑐)>0, then 𝑓(π‘₯) has a local minimum at π‘₯=𝑐;
  • if 𝑓′′(𝑐)=0, then 𝑓(π‘₯) may have a point of inflection at π‘₯=𝑐.

In this explainer, we are interested in maximum and minimum values of functions and their applications to real-world problems. Thus, we will ignore points with 𝑓′′(𝑐)=0 as the second derivative test says nothing about the point π‘₯=𝑐, a possible inflection point, with respect to optimization.

For these problems, it can be useful to begin by looking at how to apply the process in a specific example with the volume of a hot-air balloon. In particular, we will apply optimization to find the maximum volume, given its expansion function over time.

Example 1: Finding the Maximum Value of a Function Involving a Rational Function Using the Quotient Rule

Given that the volume of a hot-air balloon grows according to the relation 𝑓(𝑑)=7000𝑑𝑑+49+4000, where the time is measured in hours, determine its maximum volume.

Answer

In this example, we want to find the maximum volume of a hot-air balloon, as depicted in the picture, which grows according to a particular relation, 𝑓(𝑑), where 𝑑 is the time measured in hours.

In order to do this, we must find the critical points of the function 𝑓(𝑑), by taking the first derivative with respect to time, 𝑑, setting the derivative equal to zero, and solving for 𝑑. We can then determine the nature of the critical points using the second derivative test.

Since our relation involves a rational function, in order to differentiate, we recall the quotient rule for differentiable functions 𝑒(𝑑) and 𝑣(𝑑): 𝑒𝑣′=π‘£π‘’β€²βˆ’π‘’π‘£β€²π‘£.

Using this with 𝑒(𝑑)=7000𝑑 and 𝑣(𝑑)=𝑑+49, we can find the first derivative of the function 𝑓(𝑑) as 𝑓′(𝑑)=7000ο€Ή49βˆ’π‘‘ο…(𝑑+49)=0.

Now, solving for 𝑑, we find that 49βˆ’π‘‘=0 and hence 𝑑=Β±7. However, only nonnegative values of 𝑑 are valid since time is always nonnegative (𝑑β‰₯0); hence, we have the only critical point 𝑑=7, and we can ignore the solution with 𝑑=βˆ’7.

We can determine the nature of this critical point using the second derivative test. Applying the quotient rule again with 𝑒(𝑑)=7000ο€Ή49βˆ’π‘‘ο…οŠ¨ and 𝑣(𝑑)=𝑑+49ο…οŠ¨οŠ¨, we find the second derivative of the volume function 𝑓(𝑑) as 𝑓′′(𝑑)=14000π‘‘βˆ’2058000𝑑(𝑑+49).

Substituting the critical point 𝑑=7 into this, we find 𝑓′′(7)=14000Γ—7βˆ’2058000Γ—7(7+49)=βˆ’50049.

Thus, since 𝑓′′(7)<0, by the second derivative test, the critical point 𝑑=7 is a local maximum of the volume function 𝑓(𝑑) of the hot-air balloon. At this critical point, the volume is given by 𝑓(7)=7000Γ—77+49+4000=500+4000=4500.

The maximum volume of the hot-air balloon is therefore 4β€Žβ€‰β€Ž500 cubic units.

Let’s consider a further real-world application of optimization to maximize the profit for a particular company. Suppose a company has a nonlinear relationship between its monthly advertising budget, Β£π‘₯ thousand, and profit, £𝑦 thousand, given by the cubic function 𝑦(π‘₯)=0.05π‘₯βˆ’1.18π‘₯+7.43π‘₯βˆ’7.35.

The company’s monthly advertising budget is never less than Β£4β€Žβ€‰β€Ž000 but cannot exceed Β£14β€Žβ€‰β€Ž000; in other words, we have 4≀π‘₯≀14. To determine the maximum profit, we begin by finding the first derivative of this function with respect to π‘₯ and setting it equal to zero: 𝑦′(π‘₯)=𝑦π‘₯=0.15π‘₯βˆ’2.36π‘₯+7.43=0.dd

Solving for π‘₯, using the quadratic formula or otherwise, we find the critical points π‘₯=4.35225,π‘₯=11.3811(6).sf

Both of these critical points lie within the range of possible π‘₯-values, 4≀π‘₯≀14. Now, to determine the nature of these points, we use the second derivative test. The second derivative of the profit function 𝑦(π‘₯) is given by 𝑦′′(π‘₯)=0.3π‘₯βˆ’2.36.

Now, substituting our critical values into the second derivative, we find 𝑦′′(4.3522)=0.3Γ—4.35225βˆ’2.36=βˆ’1.0543…,𝑦′′(11.3811)=0.3Γ—11.3811βˆ’2.36=1.0543….

Therefore, since 𝑦′′(4.3522)<0 and 𝑦′′(11.3811)>0, by the second derivative test, π‘₯=4.35225 is a local maximum and π‘₯=11.3811 is a local minimum of the profit function 𝑦(π‘₯). This is also shown in the plot of the profit function below.

Given the constraint that the advertising budget is between Β£4β€Žβ€‰β€Ž000 and Β£14β€Žβ€‰β€Ž000 (4≀π‘₯≀14), by substituting π‘₯=4.35225 into the profit function 𝑦, we find that the maximum profit the company can make during a month is 𝑦(4.35225)=0.05(4.35225)βˆ’1.18(4.35225)+7.43(4.35225)βˆ’7.35=6.75760(6),6757.60.sforΒ£

We note that when we have a particular constraint for the input values π‘₯, we should also check the value of the function at the edges of the constraint as these could give a higher or lower value compared to the local maximum or minimum respectively. For this problem, it is possible that maximum profit may be at the edges of the constraint values, but this is not the case here as shown on the plot. We can also verify this by substituting the values π‘₯=4 and π‘₯=14 into the profit function: 𝑦(4)=0.05(4)βˆ’1.18(4)+7.43(4)βˆ’7.35=6.69,𝑦(14)=0.05(14)βˆ’1.18(14)+7.43(14)βˆ’7.35=2.59.

These give lower values than we got from the local maximum with 𝑦(4.35225). Thus, π‘₯=4.35225 is indeed a local maximum of the profit function given the constraint 4≀π‘₯≀14.

In our next example, we will apply optimization to find the maximum current in an electrical circuit with an alternating current.

Example 2: Finding the Maximum Value of an Alternating Current given Its Expression with Time Using Differentiation

In a circuit of an alternating current, the current 𝐼, measured in amperes, at any moment 𝑑, measured in seconds, is given by the relation 𝐼=16𝑑+19𝑑sincos. What is the maximum value of the current in this circuit rounded to two decimal places?

Answer

In this example, we want to find the maximum value of an alternating current, as depicted in the picture, given its expression as function of time, 𝐼(𝑑).

To do this, we must find the critical points of the current 𝐼, by taking the first derivative, setting it equal to zero, and solving for 𝑑. We can then determine the nature of the critical points using the second derivative test.

Taking the first derivative for the alternating current 𝐼 with respect to 𝑑 and setting it equal to zero gives 𝐼′(𝑑)=16π‘‘βˆ’19𝑑=0.cossin

In order to solve for 𝑑, we can rearrange this expression to obtain tan𝑑=1619, and hence the critical points, in radians, from the general solution for π‘›βˆˆβ„€ are 𝑑=ο€Ό1619+π‘›πœ‹=0.6998…+π‘›πœ‹.tan

Since the trigonometric functions are periodic, we only need to consider the critical points for 𝑛=0 and 𝑛=1, corresponding to 𝑑=0.6998… and 𝑑=3.8414…, as the current 𝐼 will alternate between the values evaluated at these critical points, the maximum and minimum.

We can determine the nature of these critical points using the second derivative test. The second derivative of the current function 𝐼(𝑑) is given by 𝐼′′(𝑑)=βˆ’16π‘‘βˆ’19𝑑.sincos

Substituting the critical points 𝑑=0.6998… and 𝑑=3.8414… into the second derivative gives 𝐼′′(0.6998…)=βˆ’16(0.6998…)βˆ’19(0.6998…)=βˆ’24.8394847…,𝐼′′(3.8414…)=βˆ’16(3.8414…)βˆ’19(3.8414…)=24.8394….sincossincos

Thus, since 𝐼′′(0.6998…)<0 and 𝐼′′(3.8414…>0), by the second derivative test, 𝑑=0.6998… is a local maximum and 𝑑=3.8414… is a local minimum of the function 𝐼(𝑑) of the current.

For the critical points given by the general solution, 𝑑=0.6998…+π‘›πœ‹, we have a local maximum for even 𝑛 and a local minimum for odd 𝑛.

The maximum current can be found by substituting 𝑑=0.6998…: 𝐼(0.6998…)=16(0.6998…)+24(0.6998…)=24.8394….sincos

Therefore, the maximum value of the current rounded to two decimal places is 24.84 A.

So far, for the examples we have considered, we have been given the particular function to find the optimum values for. However, in many real-world examples, we have to derive this function first by the information given or by physical considerations.

Let’s consider an example where we must first derive the function that we want to optimize. An open box is to be constructed from a rectangular piece of card measuring 64 cm by 24 cm. A square of side length π‘₯ cm is to be cut out of each corner so that the box can be made by folding, as shown in the diagrams below. The resulting box is a rectangular prism whose base measures 64βˆ’2π‘₯ cm by 24βˆ’2π‘₯ cm and whose height is π‘₯ cm.

Recall that the formula for the volume of a rectangular box with side lengths π‘Ž and 𝑏 and height β„Ž is 𝑉=π‘Žπ‘β„Ž in cubic units. In our case, the side lengths are π‘Ž=64βˆ’2π‘₯ and 𝑏=24βˆ’2π‘₯ and the height is β„Ž=π‘₯, so the volume of the box is 𝑉(π‘₯)=(64βˆ’2π‘₯)(24βˆ’2π‘₯)π‘₯=4π‘₯βˆ’176π‘₯+1536π‘₯.

In order to find the maximum value for volume 𝑉, we find the critical points of 𝑉 and identify the maxima. Taking the first derivative and setting it equal to zero, 𝑉′(π‘₯)=12π‘₯βˆ’352π‘₯+1536=0.

Solving for π‘₯, using the quadratic formula or otherwise, we find π‘₯=24,π‘₯=163.cmcm

Now we take the second derivative of the volume function 𝑉(π‘₯) in order to classify these critical points: 𝑉′′(π‘₯)=24π‘₯βˆ’352.

Substituting the critical points into the second derivative 𝑉′′(π‘₯), we find 𝑉′′(24)=24Γ—24βˆ’352=224,𝑉′′163=24Γ—163βˆ’352=βˆ’224.

Therefore since 𝑉′′(24)>0 and 𝑉′′163<0, by the second derivative test, π‘₯=24 is a local minimum and π‘₯=163 is a local maximum of the volume 𝑉(π‘₯) of the open box. The maximum volume of the open box can be found by substituting π‘₯=163: 𝑉163=4ο€Ό163οˆβˆ’176ο€Ό163+1536ο€Ό163=10240027=3792.59(6).cmsf

So far, the problems we have encountered have involved optimizing a particular function with one variable and without any constraints on the function, but many real-world problems often involve more than one variable and have a given constraint relating the independent variables. The optimization problems we want to solve are in the form maximizeorminimizeconstraintβˆΆπ‘“(π‘₯,𝑦),βˆΆπœ†(π‘₯,𝑦)=𝐢.

The constraint πœ†(π‘₯,𝑦)=𝐢 can often be used to rewrite the function 𝑓(π‘₯,𝑦) as a function of one variable only, after which the same steps apply when optimizing a one-variable function similar to the examples above.

Let’s look at how we can apply this in a real-world example. Consider a juice carton that has a capacity of 1β€Žβ€‰β€Ž000 cm3; the design of the carton is that of a closed rectangular prism (rectangular box) where the base measures π‘₯ cm by 2π‘₯ cm and height measures β„Ž cm.

The surface area of the carton, 𝐴 cm2, and volume, 𝑉 cm3, are given by 𝐴=2ο€Ή2π‘₯+π‘₯β„Ž+2π‘₯β„Žο…=4π‘₯+6π‘₯β„Ž,𝑉=π‘₯(2π‘₯)β„Ž=2π‘₯β„Ž.

Suppose we want to minimize the surface area of the carton given the constraint from the volume 𝑉=1000cm. The optimization problem we want to solve is then given by minimizeconstraint∢𝐴=4π‘₯+6π‘₯β„Ž,βˆΆπ‘‰=2π‘₯β„Ž=1000.

We can rewrite the constraint as β„Ž=10002π‘₯=500π‘₯,

expressing the height β„Ž in terms of π‘₯. We can then rewrite the surface area 𝐴 as a function of π‘₯ only: 𝐴(π‘₯)=4π‘₯+6π‘₯β„Ž=4π‘₯+6π‘₯ο€Ό500π‘₯=4π‘₯+3000π‘₯.

In order to find the minimum surface area, we begin by taking the first derivative of this function and setting it equal to zero: 𝐴′(π‘₯)=8π‘₯βˆ’3000π‘₯=0.

Solving for π‘₯, we find the critical point as π‘₯=ο„ž30008=5√3=7.21125(6).cmsf

To determine the nature of this critical point, we substitute it into the second derivative of the surface area and apply the second derivative test. The second derivative of the area function 𝐴(π‘₯) is given by 𝐴′′(π‘₯)=8+6000π‘₯.

Substituting the critical point π‘₯=5√3 into the second derivative gives 𝐴′′5√3=8+6000ο€»5√3.

Thus, since 𝐴′′5√3>0, by the second derivative test, the critical point π‘₯=5√3 is a local minimum of the surface area 𝐴(π‘₯). The minimal surface area can be found by substituting π‘₯=5√3: 𝐴5√3=4ο€»5√3+3000ο€»5√3=300√9=624.025(6).cmsf

Now let’s look at a few examples to practice and deepen our understanding of solving optimization problems and how they apply to real-world problems. In the first example, we want to find two numbers with the minimum sum of their squares, given their sum. This may seem abstract, but it can also represent a real-world scenario, as we shall see.

Example 3: Finding Two Numbers with the Minimum Sum of Their Squares given Their Sum Using Differentiation

Find two numbers whose sum is 156 and the sum of whose squares is the least possible.

Answer

In this example, we want to find two numbers with a given sum whose sum of squares is minimal.

Let π‘₯ and 𝑦 denote the two numbers whose sum is 156. The optimization problem we want to solve is then given by minimizeconstraintβˆΆπ‘†=π‘₯+𝑦,∢π‘₯+𝑦=156.

Subtracting π‘₯ from both sides of the constraint, we have 𝑦=156βˆ’π‘₯. This allows us to rewrite the sum 𝑆 as a function of π‘₯ only: 𝑆(π‘₯)=π‘₯+(156βˆ’π‘₯)=π‘₯+ο€Ήπ‘₯βˆ’312π‘₯+24336=2π‘₯βˆ’312π‘₯+24336.

In order to determine the critical points, we take the first derivative of this function with respect to π‘₯ and set it equal to zero: 𝑆′(π‘₯)=𝑆π‘₯=4π‘₯βˆ’312=0.dd

Solving for π‘₯ then gives us the critical point as π‘₯=78. To determine the nature of this critical point, we can use the second derivative test. The second derivative of 𝑆(π‘₯) is given by 𝑆′′(π‘₯)=4.

Since the value of the second derivative is positive for all values of π‘₯ and thus 𝑆′′(78)>0, by the second derivative test, the critical point π‘₯=78 is a local minimum of the function 𝑆(π‘₯). We can now find the second number, 𝑦, by substituting π‘₯=78 into the constraint: 𝑦=156βˆ’π‘₯=156βˆ’78=78.

Therefore, the two numbers whose sum is 156 and whose sum of squares is the least possible are 78,78.

Although this example deals with an abstract problem, it could represent a mathematical model of a real-world scenario. For example, if we had to place two square shields side by side across a 156 cm gap, we might want to do that in such a way as to minimize the total area of the shields and hence the amount of material needed to make them. We could let π‘₯ be the side length (in centimetres) of the first shield and 𝑦 be the side length (in centimetres) of the second shield. We could have one small shield and one large shield, or maybe they could be roughly the same size, but, either way, to fill the 156 cm gap, the sum of π‘₯ and 𝑦 would need to be equal to 156. For this scenario, we have found from the last example that the two squares which minimize the sum of their areas would be equal and both have lengths of 78 cm.

Now, let’s look at a similar example where we use optimization to maximize the area of a rectangle, given its perimeter. We will determine the maximum area and the dimensions of the rectangle.

Example 4: Finding the Maximum Area of a Rectangular Region given Its Perimeter Using Differentiation

Find the maximum area of a rectangular piece of land that can be surrounded by a fence that is 12 metres long.

Answer

In this example, we want to determine the maximum area of a rectangular piece of land, with a given perimeter.

Let 𝑙 and 𝑀 denote the length and width of the rectangle, respectively, as shown in the diagram; the perimeter of the rectangle is given by 𝑃=2𝑙+2𝑀 which is equivalent to the length of the fence in this problem.

The optimization problem we want to solve is given by maximizeconstraint∢𝐴=𝑙𝑀,∢2𝑙+2𝑀=12.

Now, making 𝑀 the subject of the constraint, we have 𝑀=6βˆ’π‘™.

Using this, we can rewrite 𝐴 as a function of 𝑙 only: 𝐴(𝑙)=𝑙(6βˆ’π‘™)=6π‘™βˆ’π‘™.

Now we can determine the critical points of the area 𝐴 by taking the first derivative of this function and setting it equal to zero: 𝐴′(𝑙)=6βˆ’2𝑙=0.

Solving for 𝑙 gives us the critical point 𝑙=3. To determine the nature of this critical point, we use the second derivative test. The second derivative of the area function 𝐴(𝑙) is given by 𝐴′′(𝑙)=βˆ’2.

Since the value of the second derivative is negative for all values of π‘₯ and thus 𝐴′′(3)<0, by the second derivative test, the critical point 𝑙=3 is a local maximum of the area function 𝐴(𝑙). At this critical point, the area is given by 𝐴(3)=6(3)βˆ’3=9.

We also note that we can determine the width 𝑀 of the rectangle from the constraint as 𝑀=6βˆ’π‘™=6βˆ’3=3, and we find that the area is maximized if the perimeter is a square.

Thus, the maximum area of a rectangular piece of land surrounded by a fence 12 m long is given by 9 m2.

This problem is similar to the abstract one where we have to determine two numbers with a given sum π‘₯+𝑦 that give the maximum product π‘₯𝑦. Abstract optimization problems can often be used to solve real-world problems in the right context.

Now, let’s look at an example where we have to find the dimensions of a rectangular cross section of a block of wood cut from a cylindrical log by maximizing the resistance of the block.

Example 5: Finding the Dimensions of a Rectangular Body Using Differentiation

The rectangular cross section of a block of wood is cut from a cylindrical log of diameter 67 cm. The resistance of this block is proportional to its width and the square of its length. What dimensions give the maximum resistance?

Answer

In this example, we want to find the dimensions of the rectangular cross section, a block of wood cut from a cylindrical log of a given diameter, which maximizes its resistance. Let’s denote the width of the block as 𝑀 and the length as 𝑙, as depicted in the diagram.

The resistance, which we can denote as 𝑅, is a measure of the strength of the rectangular block and it is always positive. Since the resistance is proportional to its width 𝑀 and the square of its length 𝑙, we have the relation, 𝑅=π‘˜π‘€π‘™, for some π‘˜βˆˆβ„οŠ°. The width and length are also related to the diameter of the log by using the Pythagorean theorem on the right triangle as 𝑀+𝑙=67. The optimization problem we want to solve is then given by maximizeconstraintβˆΆπ‘…=π‘˜π‘€π‘™,βˆΆπ‘€+𝑙=4489.

We can rearrange the constraint as 𝑙=4489βˆ’π‘€οŠ¨οŠ¨ and using this we can rewrite the resistance 𝑅 as a function of 𝑀 only: 𝑅(𝑀)=π‘˜π‘€ο€Ή4489βˆ’π‘€ο…=4489π‘˜π‘€βˆ’π‘˜π‘€.

Now we can find the critical points by taking the first derivative of this function and setting it equal to zero: 𝑅′(𝑀)=4489π‘˜βˆ’3π‘˜π‘€=0.

Solving this for 𝑀 gives us 𝑀=Β±67√33.

Since only nonnegative values of the width are valid, we can ignore the negative solution and the only critical point is 𝑀=67√33.

To determine the nature of this critical point, we use the second derivative test. The second derivative of the resistance function 𝑅(𝑀) is given by 𝑅′′(𝑀)=βˆ’6π‘˜π‘€.

Substituting the critical value 𝑀=67√33 into the second derivative gives 𝑅′′67√33=βˆ’6π‘˜ο€Ώ67√33.

Since π‘˜>0 and thus 𝑅′′67√33<0, by the second derivative test, the critical point 𝑀=67√33 is a local maximum of the resistance function 𝑅(𝑀). The length of the rectangular block can also be found from the constraint by substituting the critical point 𝑀=67√33 as follows: 𝑙=√4489βˆ’π‘€=ο„‘ο„£ο„£ο„ 4489βˆ’ο€Ώ67√33=ο„ž4489βˆ’44893=ο„ž89783=67√63.

Thus, the dimensions of the rectangular cross section that give the maximum resistance are 67√33 cm, 67√63 cm.

In the next example, we will find the maximum area of a rectangular shaped playground which ends in two semicircles, with a given perimeter.

Example 6: Finding the Maximum Area of a Region Composed of a Rectangle Ending in Two Semicircles given the Perimeter of the Region Using Differentiation

A rectangular-shaped playground ends in two semicircles. Given that the perimeter of the playground is 594 m, determine its maximum area.

Answer

In this example, we want to find the maximum area of a playground, composed of a rectangle ending in two semicircles, with a given perimeter.

Let’s denote the width of the inner rectangle as π‘₯, which is also the diameter of the semicircles, and the length of the rectangle as 𝑦. Since there are two semicircles, one on each end of the rectangle, the area and perimeter of the two ends combined are that of a whole circle, of radius π‘Ÿ=π‘₯2, given by 𝐴=πœ‹π‘Ÿ=πœ‹ο€»π‘₯2=πœ‹π‘₯4,𝑃=2πœ‹π‘Ÿ=2πœ‹ο€»π‘₯2=πœ‹π‘₯.circlecircle

The area of the playground, 𝐴pg, will be the sum of the area of the inner rectangle and the total area of the two semicircles (or whole circle): 𝐴=π‘₯𝑦+πœ‹π‘₯4.pg

The perimeter is the length of the outline of the playground, which includes the length of the inner rectangle twice and the perimeter or circumference of the two semicircles (or whole circle): 𝑃=2𝑦+πœ‹π‘₯.pg

Since we are given that the perimeter of the playground is 594 m, the optimization problem we want to solve is then given by maximizeconstraint∢𝐴=π‘₯𝑦+πœ‹π‘₯4,βˆΆπ‘ƒ=2𝑦+πœ‹π‘₯=594.pgpg

Now, making 𝑦 the subject of the constraint, we have 𝑦=297βˆ’πœ‹π‘₯2.

Using this, we can rewrite the total area of the playground 𝐴pg as a function of π‘₯ only: 𝐴(π‘₯)=π‘₯ο€»297βˆ’πœ‹π‘₯2+πœ‹π‘₯4=297π‘₯βˆ’πœ‹π‘₯4.pg

Now, we can find the critical points by taking the first derivative of this function and setting it equal to zero: 𝐴′(π‘₯)=297βˆ’πœ‹π‘₯2=0.pg

Solving for π‘₯ gives π‘₯=594πœ‹.

To determine the nature of this critical point, we use the second derivative test. The second derivative of the area function 𝐴(π‘₯)pg is given by 𝐴′′(π‘₯)=βˆ’πœ‹2.pg

Since the value of the second derivative is negative for all values of π‘₯ and thus 𝐴′′594πœ‹οˆ<0, by the second derivative test, the critical point π‘₯=594πœ‹ is a local maximum of the area function 𝐴(π‘₯)pg of the playground. At this critical point, the area is given by 𝐴594πœ‹οˆ=297ο€Ό594πœ‹οˆ+πœ‹4ο€Ό594πœ‹οˆ=176418πœ‹βˆ’3528364πœ‹=88209πœ‹.pg

Thus, the maximum area of the playground with a perimeter of 594 m is 88209πœ‹ m2.

Again, the perimeter in this example could represent the wire fencing placed around the edge of the playground.

Now, let’s consider an example where we find the radius of a circular sector with a given area that minimizes the perimeter.

Example 7: Finding the Radius of the Circular Sector with a Given Area That Has the Minimum Perimeter Using Differentiation

A sector of a circle has area 16 cm2. Find the radius π‘Ÿ that minimizes its perimeter, and then determine the corresponding angle πœƒ in radians.

Answer

In this example, we want to find the radius of a circular sector, as shown in the diagram, that has the minimal perimeter for a given area and the corresponding angle πœƒ.

The area, 𝐴sector, and perimeter, 𝑃sector, of a sector with radius π‘Ÿ and angle πœƒ are given by 𝐴=12π‘Ÿπœƒ,𝑃=π‘Ÿπœƒ+2π‘Ÿ,sectorsector where π‘Ÿπœƒ is the arc length of the segment. The optimization problem we want to solve is then minimizeconstraintβˆΆπ‘ƒ=π‘Ÿπœƒ+2π‘Ÿ,∢12π‘Ÿπœƒ=16.sector

Now making πœƒ the subject in the constraint, we have πœƒ=32π‘Ÿ.

Using this, we can rewrite the sector perimeter 𝑃sector as a function of π‘Ÿ only: 𝑃(π‘Ÿ)=π‘Ÿο€Ό32π‘Ÿοˆ+2π‘Ÿ=32π‘Ÿ+2π‘Ÿ.sector

Now, we can determine the critical points by taking the first derivative of this function and setting it equal to zero: 𝑃′(π‘Ÿ)=βˆ’32π‘Ÿ+2=0.sector

Solving this for π‘Ÿ gives us π‘Ÿ=Β±4. However, only nonnegative values of π‘Ÿ are valid since the radius is always nonnegative (π‘Ÿβ‰₯0) and we have the only critical point π‘Ÿ=4. To determine the nature of this critical point, we use the second derivative test. The second derivative of the perimeter function 𝑃(π‘Ÿ)sector is given by 𝑃′′(π‘Ÿ)=64π‘Ÿ.sector

Substituting the critical point π‘Ÿ=4 into the second derivative, we have 𝑃′′(4)=644.sector

Thus, since 𝑃′′(4)>0sector, by the second derivative test, the critical point π‘Ÿ=4 is a local minimum of the perimeter function 𝑃(π‘Ÿ)sector. We can find the value of the angle πœƒ by substituting π‘Ÿ=4 into the constraint: πœƒ=32π‘Ÿ=324=2.

Thus, we find that the radius π‘Ÿ and corresponding angle πœƒ, in radians, that minimize the perimeter are given by π‘Ÿ=4,πœƒ=2.cmrad

The sector in this example can represent a slice of pizza, where you want to find the radius and angle of the slice that minimizes its perimeter, with a fixed area.

In the next example, we will find the maximum volume of a cylinder, with a given surface area.

Example 8: Finding the Maximum Volume of a Cylinder with a Given Surface Area

What is the maximum volume of a right circular cylinder with surface area 24πœ‹ cm2? Give your answer in terms of πœ‹.

Answer

In this example, we want to find the maximum volume of a cylinder with a fixed given surface area, as shown in the diagram.

The surface area and volume of a right circular cylinder with radius π‘Ÿ and height β„Ž are given by 𝐴=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ,𝑉=πœ‹π‘Ÿβ„Ž.cylindercylinder

The optimization problem we want to solve is given by maximizeconstraintβˆΆπ‘‰=πœ‹π‘Ÿβ„Ž,∢2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ=24πœ‹.cylinder

Now, making β„Ž the subject in the constraint, we have β„Ž=12βˆ’π‘Ÿπ‘Ÿ.

Using this, we can rewrite the cylinder volume 𝑉cylinder as a function of π‘Ÿ only: 𝑉(π‘Ÿ)=πœ‹π‘Ÿο€Ύ12βˆ’π‘Ÿπ‘ŸοŠ=πœ‹π‘Ÿο€Ή12βˆ’π‘Ÿο….cylinder

Now, we can determine the critical points by taking the first derivative of this function and setting it equal to zero: 𝑉′(π‘Ÿ)=12πœ‹βˆ’3πœ‹π‘Ÿ=0.cylinder

Solving this for π‘Ÿ gives us π‘Ÿ=Β±2. However, only nonnegative values of π‘Ÿ are valid since the radius is always nonnegative (π‘Ÿβ‰₯0) and we have the only critical point π‘Ÿ=2. We can determine the nature of this critical point using the second derivative test. The second derivative of the volume function 𝑉(π‘Ÿ)cylinder is given by 𝑉′′(π‘Ÿ)=βˆ’6πœ‹π‘Ÿ.cylinder

Substituting the critical point π‘Ÿ=2 into the second derivative, we have 𝑉′′(2)=βˆ’12πœ‹.cylinder

Thus, since 𝑉′′(2)<0cylinder, by the second derivative test, the critical point π‘Ÿ=2 is a local maximum of the volume function 𝑉(π‘Ÿ)cylinder. At this critical point, the volume is given by 𝑉(2)=πœ‹Γ—2Γ—ο€Ή12βˆ’2=16πœ‹.cylinder

The maximum volume of the cylinder with a fixed surface area is therefore 16πœ‹ cm3.

The cylinder in this example can depict a soda can or a tin containing food, and you want to maximize the volume of the cylinder from the amount of material you have available, such as aluminum, corresponding to the surface area.

In the next example, we will demonstrate how to use the properties of a sphere and cylinder to maximize the sum of their volumes.

Example 9: Finding the Radius of a Sphere and a Cylinder That Maximizes the Sum of Their Volumes given the Sum of Their Surface Areas Using Differentiation

Given that the sum of the surface areas of a sphere and a right circular cylinder is 1000πœ‹ cm2 and their radii are equal, find the radius of the sphere that makes the sum of their volume at its maximum value.

Answer

In this example, we want to find the radius that maximizes the sum of the volumes of a sphere and cylinder, as depicted in the diagram, given a fixed sum of their surface areas.

The surface area and volume of a sphere and right circular cylinder of radius π‘Ÿ and height β„Ž are given by 𝐴=4πœ‹π‘Ÿ,𝑉=43πœ‹π‘Ÿ,𝐴=2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ,𝑉=πœ‹π‘Ÿβ„Ž.spherespherecylindercylinder

We are given the constraint that the sum of their surface areas is given by 1000πœ‹ cm2, so we have 𝐴=𝐴+𝐴=4πœ‹π‘Ÿ+2πœ‹π‘Ÿβ„Ž+2πœ‹π‘Ÿ=1000πœ‹.totalspherecylinder

This expression simplifies to the constraint 3π‘Ÿ+π‘Ÿβ„Ž=500. The optimization problem we want to solve is then given by maximizeconstraintβˆΆπ‘‰=𝑉+𝑉=43πœ‹π‘Ÿ+πœ‹π‘Ÿβ„Ž,∢3π‘Ÿ+π‘Ÿβ„Ž=500.totalspherecylinder

Now, making β„Ž the subject of the constraint, we have β„Ž=500βˆ’3π‘Ÿπ‘Ÿ.

Using this, we can rewrite the total volume 𝑉total as a function of π‘Ÿ only: 𝑉(π‘Ÿ)=43πœ‹π‘Ÿ+πœ‹π‘Ÿο€Ύ500βˆ’3π‘Ÿπ‘ŸοŠ=43πœ‹π‘Ÿ+500πœ‹π‘Ÿβˆ’3πœ‹π‘Ÿ.total

Now we can find the critical points by taking the first derivative of this function and setting it equal to zero: 𝑉′(π‘Ÿ)=500πœ‹βˆ’5πœ‹π‘Ÿ=0.total

Solving this for π‘Ÿ gives us π‘Ÿ=Β±10. However, only nonnegative values of π‘Ÿ are valid since the radius is always nonnegative (π‘Ÿβ‰₯0) and we have the only critical point π‘Ÿ=10. To determine the nature of this critical point, we use the second derivative test. The second derivative of the volume function 𝑉(π‘Ÿ)total is given by 𝑉′′(π‘Ÿ)=βˆ’10πœ‹π‘Ÿ.total

Substituting the critical value π‘Ÿ=10 into the second derivative, we have 𝑉′′(10)=βˆ’100πœ‹.total

Thus, since 𝑉′′(10)<0total, by the second derivative test, the critical point π‘Ÿ=10 is a local maximum of the total volume function 𝑉(π‘Ÿ)total. Therefore, the radius of the sphere that makes the sum of their volume at its maximum value is 10 cm.

This example could represent a garden feature or decoration, a wooden sphere or a dome light, placed on a cylindrical support.

In the next example, we will look at how to find the points on a curve that are closest to a given point using optimization.

Example 10: Finding the Points on a Given Curve Closest to a Given Point Using Differentiation

Find the points on the curve 𝑦=2π‘₯+21 that are closest to the point (βˆ’6,0).

Answer

In this example, we want to find the points on a particular curve, as shown in the diagram below, with minimal distance to a given point.

By the Pythagorean theorem, the distance between two points (π‘₯,𝑦) and (π‘Ž,𝑏) is given by 𝐿=(π‘₯βˆ’π‘Ž)+(π‘¦βˆ’π‘).

Thus, for any points (π‘₯,𝑦) on the curve and for (π‘Ž,𝑏)=(βˆ’6,0), this distance is given by 𝐿=(π‘₯+6)+𝑦.

The optimization problem we want to solve is therefore minimizeconstraint∢𝐿=(π‘₯+6)+𝑦,βˆΆπ‘¦=2π‘₯+21.

Using the constraint given by the equation of the curve and substituting for π‘¦οŠ¨, we can write 𝐿 as a function for π‘₯ only: 𝐿(π‘₯)=(π‘₯+6)+(2π‘₯+21)=√(π‘₯+12π‘₯+36)+(2π‘₯+21)=√π‘₯+14π‘₯+57.

We can now determine the critical points of this function by taking the first derivative and setting it equal to zero. Using the chain rule for two differentiable functions 𝑓 and 𝑔, (𝑓(𝑔(π‘₯)))β€²=𝑔′(π‘₯)𝑓′(𝑔(π‘₯)), with 𝑓(π‘₯)=√π‘₯ and 𝑔(π‘₯)=π‘₯+14π‘₯+57, we obtain 𝐿′(π‘₯)=π‘₯+7√π‘₯+14π‘₯+57=0.

Solving for π‘₯ gives us the critical point π‘₯=βˆ’7. We can determine the nature of this critical point using the second derivative test. Again, using the chain rule, the second derivative of the length function 𝐿(π‘₯) is given by 𝐿′′(π‘₯)=8(π‘₯+14π‘₯+57).

Substituting the critical point π‘₯=βˆ’7 into the second derivative, we have 𝐿′′(βˆ’7)=8(βˆ’7)+14(βˆ’7)+57=√24.

Thus, since 𝐿′′(βˆ’7)>0, by the second derivative test, the critical point π‘₯=βˆ’7 is a local minimum of the distance function 𝐿(π‘₯) from the curve. The 𝑦-coordinate can be found from the constraint or equation of the curve: 𝑦=2π‘₯+21=2(βˆ’7)+21=7.

Therefore, 𝑦=±√7. Thus, the points on the curve that are closest to (βˆ’6,0) are ο€»βˆ’7,√7, ο€»βˆ’7,βˆ’βˆš7.

In celestial mechanics, a parabolic trajectory is a Kepler orbit, with the eccentricity equal to 1, that is exactly on the border between an elliptical and hyperbolic orbit.

The parabolic curve 𝑦=2π‘₯+21 can represent the trajectory of an object around a massive body, such as the Sun, located at a particular point, (βˆ’6,0), shown by the red dot in the plot in the previous example. The abstract optimization problem can be used to find the points in the trajectory of an object around a massive body, at which the object is closest to the massive body.

Key Points

A real-world optimization problem can be solved in the following way:

  • Define the problem: specify the variables and the function to be optimized (minimized or maximized), subject to a constraint relating the independent variables.
  • Where appropriate, use the constraint to rewrite the function to be optimized as a function of one variable.
  • Differentiate the function to be optimized, setting the derivative to zero and solving for the variable, to determine the critical points.
  • Find the second derivative and substitute the critical points to determine their nature using the second derivative test: 𝑓′′(π‘₯)<0 at a maximum and 𝑓′′(π‘₯)>0 at a minimum.
  • The maximum or minimum is then found by substituting the appropriate critical point into the function to be optimized. The constraint can also be used to determine the other independent variables.

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