Question

Justin wants to use 188ft of fencing to,fence off the greatest possible retangular area for a garden .What dimensions should he use? What will be the area of the garden?

Answer 1

The greatest area that can be enclosed by a given amount of material will always be a square...

M=2L+2W  so we can say:

2L=M-2W

L=(M-2W)/2

area is:

A=LW  and using L from above:

A=W(M-2W)/2

A=(MW-2W^2)/2

dA/dW=(M-4W)/2 

d2A/dW2=-4

Since acceleration, d2A/dW2, is a negative constant, when velocity, dA/dW=0, it is at an absolute maximum for A(W).

dA/dW=0 only when M-4W=0, 4W=M, W=M/4

from earlier we found L=(M-2W)/2 and using W from above we get:

L=(M-M/2)/2

L=(2M-M)/4

L=M/4

So L=W=M/4, thus it is a square.

That's the proof, anyway...

188=2x+2y

94=x+y

y=94-x

A=xy and using y from above:

A=94x-x^2

dA/dx=94-2x, d2A/dx2=-2, as explained earlier, since acceleration is a negative constant, when dA/dx=0, it is at an absolute maximum for A(x).

dA/dx=0 only when 94-2x=0, 2x=94, x=47

and from earlier, y=94-x, so y=94-47

y=47

So the dimensions that produce the greatest area are:

47ft by 47ft

And thus the greatest area is:

47^2=2209 ft^2