Hi everyone

Question

I am lost on how to do this math problem:

A. You have 180 feet of fencing with which to construct a rectangular pen. What is the maximum area you can enclose?

B. Going back to part A, you decide to build your rectangular pen next to a large barn. (Assume the barn is more than 180 feet long, or at least long enough to be used as one side of your pen.) Now what is the maximum area you can enclose?

What I know is an area of a rectangle is (base)(height) but I'm lost on how to do this problem exactly.
Any help is greatly appreciated!

Answer 1

it would be nice to know how much derivative calculus you know.

a) area=LW
Perimter=180=2L+2W
or W=90-L

Area=L(90-L)=90L-L^2
Ok: non calculus approach. That equation is a parabola, with roots at L=0, or L=90. So the max must be halfway, or L=45. Then W=45. Max area=45^2
Now calculus approach:
Area=90L-L^2
dArea/dL= 90-2L=0 or L=45, and W-45

b. Area=LW
180=L+2W

Area=(18O-2W)W
Area= 180W-2W^2
zeroes at W=90, W=0, so max is again at W=45, L=90
calculus approach:
Area=(180W-2W^2
dArea/dW=180-4W=0 W=45, L=90

Answer 2

For a given perimeter, the largest area a rectangle can have is when the rectangle is a square, so

a) side = 45 ft
area = 45^2 or 2025 ft^2

b) now we need only one length, call it y ft
let the width be x ft

2x + y = 180
y = 180-2x

area = xy = x(180-2x)
= -2x^2 + 180x
this is a quadratic function whose graph opens downwards, thus we have a maximum area
the vertex will give us that.
The x of the vertex is -b/(2a)
= -180/-4 = 45

so the width is 45 ft and the
length = 90 ft
for a maximum area of 4050 ft^2

Answer 3

Thanks for the help, I just started this math course and I'm pretty sure I don't know much derivative calculus yet!