# Find largest area that can be enclosed in 3 conjoined rectangles.

A farmer has sixty meters of fence with which to build a rectangular animal run as shown for her cows, horses, and pigs. She wants each type of animal to have the same area. What is the largest number of square meters that can be enclosed?

As you can see in the diagram below there are 3 rectangles stacked side-by-side, each rectangle representing one pen for one animal. While attempting this problem, I got the length (horizontal side) of one of the pens as 0 meters and the width as 15 which equates to a grand area of 0 meters^2. I looked for side-lengths because the right lengths would lead you to the largest enclosed area. I have yet to find the solution. The answer choices (in meters) are,

1. 112.5
2. 124
3. 128.5
4. 135
5. 136.5

Brute forcing the solution is probably pointless because over half of the possible answers are decimals. Any help with this problem is appreciated.

• How can the length of one side be $0$ meters? I guess you really do mean zero meters as your "grand area" is $0$ square meters. – coffeemath Feb 14 at 22:33
• I used a faulty formula so I got an impossible answer of 0 meters in length and 15 meters in width, which adds up to 60 meters of fencing. I still haven't solved for the real length and width of one pen, which would give me the max area enclosed. – Jamlandia Feb 14 at 22:36
• It looks like you have successfully minimized the area. Now can you maximize the area? – Doug M Feb 14 at 23:02

Some hints: If the length (horizontal) of each pen is $$x$$ and vertical width is $$y$$ then the total fencing used is $$6x+4y$$ which to use up the fencing must be $$60$$ meters. Each pen has area $$xy$$ so total area enclosed is $$3xy$$ from the 3 pens.

Now you can eliminate $$y$$ from $$6x+4y=60,$$ plug that into total area and do calc max (derivative etc) to get first the $$x$$ value and then the max area.

Let $$w$$ and $$l$$ be the width and the length of the area. Then, $$4w+2l =60$$, or

$$l= 30-2w$$

The area is

$$A=wl= 30w-2w^2$$

Set $$\frac {dA}{dw}=0$$ to get the optimal width at $$w=\frac{15}2$$ and in turn the length at $$l=15$$. Thus, the largest square meters are

$$A_{max}=\frac{15}2\cdot 15= 112.5 m^2$$.