Im doing my homework and it provides barely any info. here is the question.
Mr Harrison has a very large backyard. He contacted the Clover Pool Company to have a pool installed in the shape of a rectangle. The Clover Pool Company would build a pool that has a perimeter of 72 yards. Mr. Harrison would be allowed to choose the dimensions (length and width) of the pool.He chose the dimensions that would give him a greatest area possible. What are the dimensions of the pool that would give Mr. the greatest area?
If L is the length and W is the width,
2(L+W) = 72
L+W = 36
W = 36 -L
Area = L(36 -L)= 36L- L^2
Set dA/dL = 0 = 0 and solve for L, if you know calculus. If you don't, rewrite the above equation as
Area = 18^2 - (L-18)^2
This is a maximum when the (L-18) term is zero, so L = 18.
thanks dude you rock
To find the dimensions that would give Mr. Harrison the greatest area for his pool, we can use the concept of derivatives and optimization.
Let's assume the length of the pool is L yards and the width is W yards. The perimeter of a rectangle is given by the formula P = 2L + 2W. In this case, the perimeter is given as 72 yards. So we can write the equation as:
2L + 2W = 72
To find the dimensions that maximize the area, we need to maximize the area function A(L, W), which is given by the formula A = L * W.
Now, we want to find the dimensions that maximize the area, so we need to find the maximum value of A(L, W) subject to the constraint of the given perimeter equation.
To solve this problem, we can use the method of Lagrange multipliers. We need to find the critical points of the function A(L, W) subject to the constraint equation.
Let's set up the Lagrange function as follows:
L(L, W, λ) = A(L, W) - λ(2L + 2W - 72)
To find the critical points, we need to take partial derivatives of L with respect to L, W, and λ, and set them equal to zero:
∂L/∂L = 0 (partial derivative of L with respect to L)
∂L/∂W = 0 (partial derivative of L with respect to W)
∂L/∂λ = 0 (partial derivative of L with respect to λ)
Taking the partial derivatives:
∂L/∂L = W - 2λ = 0
∂L/∂W = L - 2λ = 0
∂L/∂λ = 2L + 2W - 72 = 0
From these equations, we can solve for L and W. Solving the first two equations gives us:
W = 2λ
L = 2λ
Substituting these values into the third equation:
2(2λ) + 2(2λ) - 72 = 0
4λ + 4λ - 72 = 0
8λ = 72
λ = 9
Substituting the value of λ back into the equations for L and W:
L = 18
W = 18
So the dimensions that give Mr. Harrison the greatest area for his pool are L = 18 yards and W = 18 yards.
Keep in mind that the Lagrange multiplier method is just one way to solve this problem. There are other methods that can be used as well.