Optimize Circle and Square Area?

Question

Sorry to ask such a question, but our study group is at a loss as how to continue and our homework is due tomorrow. So here goes:

In order to receive credit we MUST use calculus techniques:

We have a piece of wire that is 100cm long and we're going to cut it into two pieces. One piece will be bent into a square and the other will be bent into a circle. Determine where the wire should be cut so that the enclosed areas will be at maximum. Note that it is possible to have the whole piece of wire go either to the square or to the circle

Answer 1

You need to come up with an expression for the total area in terms of a variable that distinguishes where the wire is cut. I'd call one hunk length x and the other hunk 100 cm -x. You should be able to write out the sum of the areas of the square and circle in terms of x.

Then you use calculus to maximize the expression (you differentiate and set the derivative to zero [and solve the resulting equation] because the derivative must be zero at a maximum).

You need to manually check what the area is for x=0 and x=100cm also because the above process won't investigate those possible maximums.

Answer 2

V = x - 2y x^2 + y^2 = a million x^2 = a million - y^2 x = sqrt(a million - y^2) V = sqrt(a million - y^2) - 2y dV/dy = (a million/2) * (-2y) / sqrt(a million - y^2) - 2 dV/dy = 0 0 = -y / sqrt(a million - y^2) - 2 2 = -y / sqrt(a million - y^2) 2 * sqrt(a million - y^2) = -y 4 * (a million - y^2) = y^2 4 - 4y^2 = y^2 4 = 5y^2 4/5 = y^2 x^2 + y^2 = a million x^2 + 4/5 = a million x^2 = a million/5 x^2 = 5/25 x = +/- sqrt(5) / 5 y = +/- 2 * sqrt(5) / 5 x - 2y There are 4 plausible values sqrt(5)/5 - 4sqrt(5)/5 sqrt(5)/5 + 4sqrt(5)/5 -sqrt(5)/5 - 4sqrt(5)/5 -sqrt(5)/5 + 4sqrt(5)/5 -3sqrt(5)/5 5sqrt(5)/5 -5sqrt(5)/5 3sqrt(5)/5 5sqrt(5)/5 => sqrt(5) the optimal cost is sqrt(5) all of us be attentive to that the diagonal of the rectangle could be the diameter of the circle. it is, it extremely is going to likely be 16m x^2 + y^2 = sixteen^2 the place x and y are the size of the rectangle A = x * y A = x * sqrt(256 - x^2) dA/dx = x * (-2x) * (a million/2) / sqrt(256 - x^2) + sqrt(256 - x^2) dA/dx = 0 0 = -x^2 / sqrt(256 - x^2) + sqrt(256 - x^2) x^2 / sqrt(256 - x^2) = sqrt(256 - x^2) x^2 = 256 - x^2 2x^2 = 256 x^2 = 128 x = 8 * sqrt(2) x^2 + y^2 = sixteen^2 128 + y^2 = 256 y^2 = 128 y = 8 * sqrt(2) A = x * y A = 8 * 8 * sqrt(2) * sqrt(2) A = sixty 4 * 2 A = 128 128 m^2

Answer 3

Let x be the length of wire that you intend to bend into a square (therefore, 100-x is the length that you intend to bend into a circle).

The area of the square will then be:
A_s(x) = (side)² = (x/4)²

And the area of the circle will be:
A_c(x) = π(radius)²
= π(circumference/(2π))²
= π((100-x)/(2π))²
= (100-x)²/(4π)

Now write a function that shows the total area of the two shapes, as a function of x:

A(x) = (x/4)² + (100-x)²/(4π)

Now find the maximum of A(x). To do that, take the derivative of A(x) and set it to zero.