Calculus Optimization Problem Help?

Question

The problem goes something like this: A landscape architect wished to enclose a rectangular garden on one side by a brick wall costing $30/ft and on the other three sides by a metal fence costing $10/ft. If the area of the garden is 188 square feet, find the dimensions of the garden that minimize the cost.

So, I know that I essentially need to minimize the perimeter.

Area = L * W
We know that Area = 188
Perimeter = 2L + 2W
L = 188/W
Using substitution I get: P = 2(188/W) +2W

Then I found the derivative in order to find the critical points: P' = -376/w^2 +2

The critical point that I found that fits inside the practical domain was 13.711 which would make both L and W equal to 13.711 but the site is telling me that my answer is wrong. I have no idea where I'm going wrong with this problem, could anyone help me out?

husoski · Accepted Answer

You're supposed to minimize cost, not perimeter.

Measure the width along the brick wall, so the brick length is W and the fence length is 2L + W.  The cost is then

C = 30W + 10(W + 2L) = 40W + 20L = 40W + 3760/W

dC/dW = 40 - 3760/W^2 = 0
W^2 = 3760/40 = 94
W = sqrt(94) . . . . only the positive solution is feasible
L = 188 / sqrt(94)

That.s about 9.7 ft for the width and 19.4 ft for the length.

? · Answer

I ended up figuring out what I did wrong 😂 thanks for all the helpful answers!

? · Answer

Area xy=188
x is the brick wall side.
Cost will be (x+2y)10+30x. This has to be minimised. Use the method Lagrange undetermined multiplier  l. Write
f(x,y) = xy +l ((x+2y)10+30x)
One need to find l. Take partial derivative wrt x and y and equate them to 0. For x you will hav
y+ l 40=0. For y you will have
x+ l 20=0, You have now
xy=188= l^2 (40*20). This gives
l=0.485. It has plus and minus sign. Select minus sign. x= -20 l=9.7 ft y=19.40 ft
This method is well suited for these problems.

D g · Answer

you have to minimize the cost not the perimeter..

the perimeter is  30 dollars a foot on one side and  10 / foot on another 3 sides

the cost is  30x +10 x+ 2(10y) = total cost

x is the length of one side 
y is the length of the other side

area = xy

188 = xy
y = 188/x

substitution into total cost

40x + 2*(y) = total
40x + 2*(188/x) = total

multiply through by x

40x^2 + 376 - total x = 0

this is a quadratic formular

you need to find the minimum of this 
40x^2 - total x + 376 = 0

80x - total = 0

this is the  local minimum max point

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Calculus Optimization Problem Help?

4 Answers