32. Answer Exercise 31 if one piece is bent into a square and ...

Question

32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.

Accepted Answer

Hello. In this video, we are told that a landscaper has 30 m of fencing to create an equilateral triangular vegetable garden and a circular flower garden. If the goal is to minimize the total area covered by the garden and flower bed, how much fencing should be used for each? So let's go ahead and draw an image. First, we have an equilateral triangle that is acting as the vegetable garden, and we can label each of these sides as S. Next, we also have a circular flower garden, and this is going to have some radius R. Now, what we want to do is we want to find an equation for the total area of these gardens. Now, the area of an equilateral triangle is going to be the square root. Of 3 divided by 4 multiplied each side squared, and the area of the circular guarded flower bed is pi R squared. So, if we wanted to create an area equation that takes the area of both of these shapes, then we will just take the sum of the areas of the two shapes, and this is going to be our area equation. Furthermore, we need to create a constraint equation to find a minimized. Minimized SNR for these gardens. Now, the constraint equation could be used as the perimeters of each shape. Now, in order to find the perimeter of the equilateral triangle, we just take the sum of all three sides which are labeled as S, or in other words, it is 3S, and the perimeter of the circle is the circumference, which is 2 pi R. Now we are told that the landscaper has 30 m of fencing total, so this perimeter is going to equal to 30. So the overall idea is that we're going to want to optimize this area function, and we are going to want to optimize it in such a way that we find S or R and show that it's a minimum. So, before we take a derivative of this area, we need to express this area as a function of one variable. One way we can do this is we can solve for either S or R in our constraint equation. Let's go ahead and solve for S in terms of R. That is going to allow us to rewrite this equation as 3 S is equal to 30 minus 2 pi R. And dividing both sides of this equation by 3 gives us 30 minus 2 pi R divided by 3. This is going to be a replacement that we will use for S in the given area. This will allow us to rewrite the area function to be A is equal to the square root of 3 divided by 4, multiplied by 30 minus 2 by R, divided by 3 squared. Plusi are squared. Now, what we're going to do is we're going to simplify this function. We can factor out a 2 from this numerator, but once we pull out this 2 from the squared quantity, we'll have to square that 2, giving us 4. So that's going to give us A is equal to the square root of 3 divided by 4, multiplied by 4, which is just the square root of 3. Then this will be multiplied by the quantity of 15 minus pi R squad, and this will be divided by 32, which is 9, and we will be adding pi R squared at the end. Now, at this step right here, what we want to do is you want to go ahead and take the derivative with respect to R, in order to start optimizing. That is going to give us DA over DR, and this is going to be the square root of 3 divided by 9, multiplied by 2, multiplied by 15 minus pi R because of the chain rule, and again by the chain rule, we need to multiply by the derivative of 15 minus pi R, which is going to be i, negative pi. Then we will add the derivative of pi R2, which will be 2 PIR. Now what we need to do is we need to simplify. By combining terms together, we get that the derivative of the area with respect to R is equal to -2 pi multiplied by the square root of 3, multiplied by 15 minus pi R, all divided by 9. Plus 2 i R. Now, what we want to do is you want to go ahead and solve for R by setting this equation equal to 0. So, if we set this equation equal to 0, notice that we want to eliminate this denominator of 9. In order to do this, we will multiply the entire area equation by 9, and that will give us -2 pi multiplied by the square root of 3, multiplied by 15 minus pi R plus 18 pi R equal to 0. And once we solve for R, we get that R is equal to 15 divided by pi plus 3 square root of 3. But what we need to do now is we need to verify that this is a minimum value. So In order to verify that this is a minimum value, let's go ahead and take the second derivative of our original derivative function. The second derivative will be defined as DA2 over DR2, and that is going to equal to 2 squared multiplied by the square root of 3 divided by 9 plus 2 pi. Notice that for whatever value we plug into this function, it is always positive. And because this is positive by the second derivative test, this derivative of AD ADR is going to be a minimum, meaning that R is verified to be a minimum value. Now we need to solve for S. But in order to solve for S, all we need to do is just take our R value we solve for, and plug it into our original constraint function. That is going to give us S is equal to 30 minus 2 pi multiplied by 15 divided by pi plus 3 square root of 3, and this will all be divided by 3, and S is going to simplify to be 30 square root of 3 divided by pi plus 3 square root of 3. Now, recall that we want to find a minim uh we want to find the areas of SNR that minimize the total area and how much fencing should be used. Well, now that we have SNR and we verified that they're minimums, in order to figure out how much fencing we need, all we have to do is just plug in our SNR back into the respective perimeters. So for the triangle. The perimeter of the triangle was 3S, and if we plug in the value of S we solved for, we get 3 multiplied by 30 multiplied by the square root of 3, divided by pi plus 3 square root of 3, and that is going to give us S is equal to 90. Square root of 3, divided by pi plus 3 square root of 3. And that is going to be our first solution. Now for the circle, we are going to plug in our value of R into the circumference. That is going to be 2 pi R, which is equal to 2 pi multiplied by 15 divided by pi plus 3 square root of 3. And once you multiply these values together, we get 30 pi. Divided by pi plus 3 square root of 3, which is going to be the solution for the circles for the circle's perimeter. And with that being said, the solution to this problem is going to be B. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

32. Answer Exercise 31 if one piece is bent into a square and the other into a circle.

Key Concepts

Perimeter and Area

Geometric Shapes

Optimization

Watch next