A rancher has 1000 feet of fencing in which to construct adjacent, equally sized rectangular pens. What dimensions should these pens have to maximize the enclosed area?
Perimeter of the pens is P = 3L + 4W = 1000 Area of the pen is A = 2LW
WE have to write one variable in terms of the other variable so I wrote rewrite the perimeter as a function of the Width:
\[ L = 100/3 - 4/3W \] I substituted L in this equation into the area of the pen and we get:
\[ 2W(100/3 - (4/3)W) \]
Simplifying the function we get:
\[ (200/3)W - (8/3)W^2 \]
Now that we have our function, now we differentiate and get our critical points i.e fâ(c) = 0
\[ f'(x) = 200/3 - (16/3)W = 0 \] Solving for W we get:
\[ W = (200/16) \] We plug W into our perimeter equation and solve for L now we get that
\[ L = (50/3) \]
Now that we have our length and width equations we can solve for the equation.
\[ A = 2 * (50/3) * (200/16) \]
Dimensions of the Fence should be L = 50/3 and W = 200/16 and the area of the fence is 416.667