# A farmer intends to fence o a rectangular pen for his pig Wilbur, using the barn as one of the sides. If the enclosed area is to be 50 square feet

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A farmer intends to fence o a rectangular pen for his pig Wilbur, using the barn as one of the sides. If the enclosed area is to be 50 square feet, what is smallest amount of fence needed, in feet?

deez nutz ha got em
Though seriously, figure out what equations you need as secondary and primary equations. Do this by reading problem to see what the question is wanting you to optimize. (Primary Optimize) (Secondary Substitution)
A=50=x·y (Secondary)
P=2x+y (Optimize) The problem is asking you for fence so Perimeter is Primary.
Find y in terms of x using Area equation and plug into Perimeter equation.
y=50/x
P=2x+50/x
Optimize (Find derivative, then set equal to zero)
P’=2-(50/x^2)
0=2-(50/x^2)
-2=-50/x^2 (Multiply both sides by x^2)
2x^2=50 (Divide both sides by two)
x^2=25 (square root both sides)
x=5
Now you have found x. From here on out you are basically in the clear. All you need to do is plug x back into the Area equation to give you a y value.
50=5y (Divide by 5)
y=10
Now with both values, plug x and y into your perimeter equation to give you your y values.
2(5)+10=P
P=20 (This is the amount of fence you will have.