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csc2iffy
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Homework Statement
We have to solve this problem using only what we learned in calculus (no linear programming)
I attached a picture to help :)
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?
The attempt at a 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)
We have to solve this problem using only what we learned in calculus (no linear programming)
I attached a picture to help :)
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?
The attempt at a 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)
