We have to solve this problem using only what we learned in calculus (no linear

Question

We have to solve this problem using only what we learned in calculus (no linear programming)
Please read through the WHOLE thing, I already attempted to solve it, but was told it's wrong

A farmer wants to build a rectangular pen. He has a barn wall 40 feet long, some or all of which must be used for all or part of one side of the pen. In other words, with f feet of of fencing material, he can build a pen with a perimeter of up to f 40 feet, and remember he isn't required to use all 40 feet.
What is the maximum possible area for the pen if 100 feet of fencing is available?

MY ATTEMPTED SOLUTION
I attempted to solve by finding the critical points of the derivative:
2x y=100 --> y=100-2x
x(100-2x)=100x-2x^2
f'(x)=100-4x
100-4x=0 --> x=25
2(25) y=100 --> y=50
A=1250

BUT I realize this is wrong because if y=50, then the side of the barn needs an extra 10 feet, but all of the fencing material has been used by the other 3 sides. Help please? (p.s. I know the answer is 30 by 40, I figured it out using LP but was told I'm not allowed to do it this way)

Explanation / Answer

Your setup looks ok, but you didn't maximize it properly under certain constraints...

the two startup equations:

2x + y = 100

A= x*(100-2x)

if you want your total perimeter to be 100, x must be inbetween 30 and 50. so from here,

A= 100x -2x^2

A' = 100 - 4x = 0

x=25

but: you must also check the boundaries of 30 and 50

A(25) = 25(100-2(25)) = 1000

A(30) = 30(100-2(30)) = 1200

A(50) = 50(100-2(50)) = 0

So...because of this, X must equal 30' and to solve for the other side

A= x*y

1200= 30'*y

y=40'

so...the dimensions are therefore 30'X40'

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