# Video: AP Calculus AB Exam 1 • Section I • Part B • Question 84

A wire of length 72 inches is cut into two equal pieces to make a circle and an equilateral triangle. If the total area of the circle and the equilateral triangle is a minimum, what is the area of the circle to the nearest square inch?

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### Video Transcript

A wire of length 72 inches is cut into two equal pieces to make a circle and an equilateral triangle. If the total area of the circle and the equilateral triangle is a minimum, what is the area of the circle to the nearest square inch?

For this question, we’ve been given information about a length of wire that has been cut to form two shapes. What this is really telling us is that the perimeter of our circle and the perimeter of our triangle will sum to have a length of exactly 72 inches. Here, we have used this information to form an equation: 𝑃 𝐶, the perimeter of the circle, plus 𝑃 𝑇, the perimeter of the triangle, is equal to 72.

The next thing we’re told is that the total area of the circle and the equilateral triangle is a minimum. This means that we’ll need to work with the total area. But in order to get here, we’ll first need to work out the individual areas of our two shapes. Given that we called the perimeter of our circle 𝑃 𝐶, we’ll call the area of our circle 𝐴 𝐶. We’ll take this as a starting point and move on to the area of our triangle later. Now, we know that the area of any circle is given by 𝜋 times its radius squared. We don’t have the radius of our circle. But we have to find its perimeter, 𝑃 𝐶.

One of the relationships we should be familiar with is that the perimeter of any circle is equal to two 𝜋 times its radius. Rearranging this, we get that 𝑃 𝐶 over two 𝜋 is equal to the radius 𝑟. We can now substitute this into our equation to get the area of our circle in terms of its perimeter. Now, with a small amount of rearranging, we arrived at the result that the area of our circle is equal to one over four 𝜋 times its perimeter squared. Let’s put this result to one side and move on to finding the area of our triangle.

We define the area of our equilateral triangle as 𝐴 𝑇. We know that the area of any triangle is half times its base times its height. Since we’re working with an equilateral triangle, we know that all its angles are equal being 60 degrees. And we also know that all of its sides are equal. Since we have three sides, we know that each of them will be equal to one-third of the perimeter, which is 𝑃 𝑇 over three. This gives us the base of our triangle in terms of its perimeter. But what about the height?

We can find this by cutting our equilateral triangle in half and noticing that we have formed a right-angled triangle, which has a hypotenuse again of 𝑃 𝑇 over three. And the height of our triangle corresponds to the side in our right-angled triangle which is opposite of a 60-degree angle. This allows us to perform a simple trigonometric calculation to find ℎ. We use the fact that sin of 60 degrees is one of the exact trigonometric ratios being the square root of three over two. And we find that our height is equal to the square root of three over six times the perimeter of the triangle.

We can now substitute this back into the equation for the area of our triangle. And we can then simplify our equation. Doing so, we find that the area of our triangle is equal to the square root of three over 36 times the perimeter 𝑃 𝑇 squared. Let’s again shift this result over to one side to make some room. Our next step will be to find an expression for the total area of our two shapes. Clearly, this will be the sum of the area of our circle and the area of our triangle.

Now since the only numerical information we’ve been given is to do with the perimeter of our two shapes, let’s express the area in terms of these perimeters. We can do this using the two equations that we have just formed for 𝐴 𝐶 and 𝐴 𝑇. Substituting in, we form an equation for the total area in terms of 𝑃 𝐶 and 𝑃 𝑇. Let’s stop and take a moment to think here. It intuitively makes sense that if we were to change the distribution of our 72 inches of wire between our two shapes, the total area of the two shapes would also change. This is mathematically reflected here in the fact that we have an equation for the total area in terms of either the two individual areas or the two individual perimeters.

Now, remember we’re working towards finding the point at which the total area of our two shapes is a minimum. What this means is that the rate of change of our total area with respect to either of the two individual areas or the two individual parameters would be zero. One of the ways that we can work towards finding our minimum is to differentiate our total area with respect to one of these four things and then further evaluate the points at which this is equal to zero. To do so, we’ll first need to express 𝐴 total just in terms of one of the four variables.

Now, recalling that we need to find the area of the circle when 𝐴 total is at a minimum, we may think to ourselves that it makes most sense to use 𝐴 𝐶 as this variable. However, as it turns out, our calculations will be much simpler if instead we use the perimeter of our circle 𝑃 𝐶. We therefore proceed down this route. Okay, how do we express 𝐴 total just in terms of 𝑃 𝐶? Luckily, we have an equation which relates 𝑃 𝐶 and 𝑃 𝑇. And this will allow us to substitute in for this 𝑃 𝑇 squared term.

By simple rearrangement of our first equation, we have that 𝑃 𝑇 is equal to 72 minus 𝑃 𝐶. We can hence perform a substitution for 𝑃 𝑇. Now that our equation is just in terms of 𝑃 𝐶, we move on to simplifying, first by multiplying out the parentheses. Here, we have squared the binomial. We then multiply by our coefficient of the square root of three over 36. And we then simplify and collect like terms. We’re now left with a quadratic equation which expresses the total area in terms of the perimeter of our circle.

Let’s again clear some room for our calculations. To work towards finding where 𝐴 total is minimized, the next thing we will be doing is differentiating 𝐴 total with respect to 𝑃 𝐶. We’ll then be evaluating where d𝐴 total by d𝑃 𝐶 is equal to zero. Performing our differentiation, we’re left with the following equation. Again, we want to find the point or points at which this rate of change is equal to zero. To solve for 𝑃 𝐶, we first add four times the square root of three to both sides of our equation. And then, we simplify to isolate 𝑃 𝐶. At this point, we can use a calculator to evaluate 𝑃 𝐶. Doing so, we find that 𝑃 𝐶 is equal to 27.129 et cetera inches.

Now, here is worth noting that we found the value for 𝑃 𝐶 for which the rate of change of the total area with respect to 𝑃 𝐶 is equal to zero. Now, ordinarily, we may think to differentiate again to find out whether this occurs at a maximum or a minimum. This is because a rate of change of zero or a horizontal tangent can occur at either a peak or a trough which correspond to a maximum or minimum, respectively. In actual fact, differentiating again is unnecessary here. And we can see why by looking back at the original equation for the total area in terms of just the perimeter of the circle.

Since we have a quadratic equation here, meaning the highest power of 𝑃 𝐶 is two and the coefficient of our 𝑃 𝐶 squared term is positive, we can deduce that the shape of our curve will be a positive parabola. Here, we see the top parabola opens upwards. The only point on this parabola with a slope or gradient of zero occurs when its 𝑦-value or 𝐴 total is minimised. This means that we don’t need to differentiate again to find out whether we have a minimum or a maximum because the parabola has no maximums.

That’s great. We found the perimeter of our circle when the total area is at a minimum. However, the question is actually asking us for the area of the circle at this point. Luckily, we’ve already formed an equation for the area of the circle in terms of its perimeter. So this step is fairly straightforward. We can directly substitute the value for 𝑃 𝐶 into our equation for the area of the circle. And when we evaluate, we find our answer is 58.567 et cetera inches squared.

Now since the question asked for our answer to the nearest square inch, our final step is to round this to 59. In doing this, we have answered our question. And we have found that when the total area of our two shapes is a minimum, the area of our circle is approximately equal to 59 inches squared.