Video: Finding the Dimensions of a Rectangle with Maximum Area given Its Perimeter Using Differentiation

Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.

03:56

Video Transcript

Find the dimensions of the rectangular corral producing the greatest enclosed area given 200 feet of fencing.

In order to solve a problem like this, where we need to find a greatest or least value, we need to use optimization. The maximum and minimum values of a function will occur when the derivative d𝑦 by dπ‘₯ is equal to zero. If the second derivative d two 𝑦 by dπ‘₯ squared is greater than naught, we have a minimum value. If d two 𝑦 by dπ‘₯ squared is less than naught, we have a maximum value. Let’s now consider the problem in this question. Let’s consider a rectangle of width π‘₯ feet and length 𝑦 feet. The perimeter of this rectangle is equal to two π‘₯ plus two 𝑦 and the area of the rectangle is equal to π‘₯ multiplied by 𝑦.

We’re told that we have 200 feet of fencing. Therefore, the perimeter must be equal to 200. Our first equation becomes two π‘₯ plus two 𝑦 is equal to 200. Dividing both sides of this equation by two gives us π‘₯ plus 𝑦 is equal to 100. Subtracting π‘₯ from both sides of the equation gives us 𝑦 is equal to 100 minus π‘₯. We can now substitute this into the second equation. The area of the rectangle is equal to π‘₯ multiplied by 100 minus π‘₯. Distributing the parentheses gives us 100π‘₯ minus π‘₯ squared. We can now begin to work out the dimensions of the greatest area by differentiating this expression.

If 𝐴 is equal to 100π‘₯ minus π‘₯ squared, then d𝐴 by dπ‘₯ is equal to 100 minus two π‘₯. We can differentiate each of the terms individually. Setting this equal to zero gives us 100 minus two π‘₯ equals zero. Adding two π‘₯ to both sides gives us two π‘₯ is equal to 100. And finally, dividing by two gives us a value of π‘₯ equal to 50. We know that 𝑦 is equal to 100 minus π‘₯. So we can substitute this value into the equation. 100 minus 50 is equal to 50. So 𝑦 is also equal to 50. Whilst it appears we have worked out the answers, it is important that we check that this value corresponds to the maximum or greatest value.

d𝐴 by dπ‘₯ was equal to 100 minus two π‘₯. We can differentiate this again to get d two 𝐴 by dπ‘₯ squared. Differentiating any constant β€” in this case, 100 β€” gives us zero. And differentiating negative two π‘₯ gives us negative two. Therefore, d two 𝐴 by dπ‘₯ squared is negative two. At this stage, we usually substitute in the values of π‘₯ into an expression. However, our answer is a constant, negative two. This is because the equation for 𝐴 was an n-shaped parabola and, therefore, just as one turning point. This is a maximum as d two 𝐴 by dπ‘₯ squared is negative. The dimensions that produced the greatest enclosed area are 50 feet by 50 feet.

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