Question

Greg wants to build a rectangular enclosure for his animals. One side of the pen will be against the barn, so he needs no fence on that side. The other three sides will be enclosed with wire fencing. If Greg has 650 feet of fencing, you can find the dimensions that maximize the area of the enclosure.

a) Let WW be the width of the enclosure (perpendicular to the barn) and let LL be the length of the enclosure (parallel to the barn). Write an function for the area AA of the enclosure in terms of WW. (HINT first write two equations with WW and LL and AA. Solve for LL in one equation and substitute for LL in the other).


A(W)=A(W)=    
b) What width WW would maximize the area?
ww =  ft
Round to nearest half foot c) What is the maximum area?
AA =  square feet
Round to nearest half foot

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Answer 1

Maximum area is always a square. Hence side of square =

650/4=162.5 feet

One side is not needed so add that length to the non-parallel side.

162.5+162.5=325 feet is the length and 162.5 feet is the width(s)

A=325*162.5

A=52812.5 square feet

Alternatively, we can solve by

P=l+2w

650=l+2w

650-2w=l

A=lw

A=(650-2w)w

A=650w-2w²

This is a parabola that opens downward so there is a maximum point.

The vertex of the parabola is (h,k) where h is the "maximizing number" and k is the maximum area.

Use the fact that h=-b/2a

h=-650/(2*[-2])

h=-650/(-4)

h=162.5 would be the length of all four sides if it were not for the barn

Therefore you have an extra 162.5 feet

Add the 162.5 feet to the opposite side(length) to get 325 feet.

You have a rectangle that is 162.5 feet by 325 feet by 162.5 feet by "the barn".

The width is 162.5 feet which maximizes the area.

A=lw

A=325*162.5

A=52812.5 square feet is the maximum area.

If we substituted 162.5 for w in the equation A=650w-2w² you would get 52812.5.

So 212.5 is the maximizing number, h, and the maximum area is k, which is 52812.5 square feet.