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# Challenge Ap Calc Problem?

Twenty feet of wire is to be used to form two figures. In each of the following cases, how much should be used for each figure so that the total enclosed area is a maximum.

a.) equilateral triangle and square

b.) square and regular pentagon

c.) regular pentagon and regular hexagon

d.) regular hexagon and circle

What can you conclude from this pattern?

[HINT: the area of a regular polygon with n sides of length x is a= n/4[cot(pi/n)]x^2

Any Help?

### 2 Answers

- 1 decade agoFavorite Answer
I just gave a hint to your other AP calculus question. The problem here is exactly the same. You have two equations, two variables.

I'll do part a) as an example.

You have 20 feet of wire and need to make an equilateral triangle and a square. That's an equation for perimeter.

P = 3x + 4y = 20

Now you need an equation for area.

A = x^2*sqrt(3)/2 + y^2

Two equations, two variables. In the perimeter equation, solve for y as a function of x, plug into the area equation to get: A = x^2*sqrt(3)/2 + (5 - 3/4x)^2

Now you have the area as a function of only one variable. Time to take the derivative in respect to x.

dA/dx = x*sqrt(3) - (3/2)*(5 - 3/4x)

Set that equal to zero and solve for x to get:

x = 1/(9/8 + sqrt(3))*(15/2)

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- osiasLv 43 years ago
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