I am stuck on another optimization problem and I can't get my answer to match the author's. I am assuming the author is correct, but there is no justification for their answer.
A farmer with 750 ft of fencing wants to enclose a rectangular area and then divide it into four pens with fencing parallel to one side of the rectangle. What is the largest possible total area of the four pens?
I know that the two formulas I need are
$2w + 5l = 750$ w for width and l for length to find the perimeter. I know it is this because there are 3 fences laid down inside a rectangle, so that gives me 5 lines of fence in one way and then 2 in the other direction.
$4lw = a$ for this I know that I have 4 rectangles so it is 4 lw.
Subsituting in w in terms of l
$2l(750-5l) = a$
$1500l-10l^2=a$
then I take the derivative
$1500 – 20l = a$
$l=75$ which gives me the wrong answer.
Best Answer
Your mistake seems to be that you're confusing the width of each pen with the width of the whole area. The formula $2w + 5l = 750$ assumes that $w$ is the width of all four pens added together, yet $4lw = a$ treats $w$ as the width of only one pen. Once you make the definition of $w$ consistent, you should get the right answer.