
Zachary K. answered 03/24/19
Zak the Math Blackbelt
Recall the formulas for perimeter and area of a rectangle. Now adjust the perimeter formula to account for the missing side. Keeping with the wording of the question it is one of the lengths that is missing. So two widths and one length must add up to the given amount of fencing, lets call the amount of fencing P.
Thus P = 2W+L.
We can use this information to solve for one of the missing variables in terms of the other. L=P-2W.
And since area of a rectangle is given by A=L*W, we can construct a formula for the area of the pen.
A=L*W, but we know L=P-2W, subbing this in we get
A=(P-2W)W
Or
A=P*W - 2W2 (A quadratic!)
A quick application of the vertex formula (-b/(2a), f(-b/(2a))) will give us the W that maximizes the area. (-b/(2a)) and the maximized Area ( f(-b/(2a)). To get the missing value of L plug the W value into the L=P-2W formula.