A rectangular field is bounded on one side by a river and on the other three sides by a fence. Additional fencing is used to divide the field into three smaller rectangles, each of equal area. 1080 feet of fencing is required. I want to find the dimensions of the large rectangle that will maximize the area.

I had the following equations and I was wondering if they are correct:

I let $Y$ denote the length of the rectangle. Then $Y=3y$. If $x$ represents the width of the three smaller rectangles, then I get the following: $$ 4x+3y = 1080,~~~\text{Area} = 3xy.$$

  • $\begingroup$ Apart from the fact that the cows won't be able to drink at the river, we can do better by using two division fences that are perpendicular to each other, the one perpendicular to the river going all the way across, and the one parallel to the river going only partway. $\endgroup$ Apr 24 '13 at 20:57

Now that I understand your "layout", yes, the formulae you created will work just fine, remembering that little "y" is the dimension of the length of a smaller rectangle, and $Y = 3y$ the dimension of the length of the large rectangle encompassing the whole.

$$ 4x+3y = 1080,~~~\text{Area} = 3xy.$$

We can then express Area as a function of $x$, first solving for $y$ in the first equation, and substituting this expression for $y$ into the Area equation:

$$4x + 3y = 1080 \iff 3y = 1080 - 4x \iff y = 360 - \frac 43 x$$

$$\text{Area}\;= 3x(360 - (4/3)x) = x(1080 - 4x) = = 1080x - 4x^2$$

To maximize area, we calculate $A'$ and set $A' = 0$ to determine the function's critical point:

$$A' = 1080 - 8x,\qquad A' = 0 \implies 1080 - 8x = 0 \implies x = 135. $$ We can easily show that the solution to $A' = 0 \iff x = 135$ ft. does in fact give the maximum value for $A$.

Now we have $x$, and will need only to compute $y = 360 - 4/3(135) = 360 - 180 = 180$ ft.

Then $Y = 3 \times 180 = 540$ ft, and so we've determined the larger rectangle's dimensions for maximized area: $ Y$ ft $ \times x$ ft $= 540 \text{ ft} \times 135 \text{ ft}$, with smaller rectangles each $180 \text{ ft} \times 135\text{ ft}$

  • $\begingroup$ I am sorry. I should have mentioned the fact that I was given a picture as well. In the picture the river is at the opposite side of the length, $y$. $\endgroup$
    – Gorg
    Apr 24 '13 at 20:42
  • $\begingroup$ @amWhy: such a nice, well written and guiding answer. =1 $\endgroup$
    – Amzoti
    Apr 25 '13 at 1:43
  • $\begingroup$ Gorg: I figured out your "picture"...let me know if my answer helps at all! ;-) $\endgroup$
    – amWhy
    Apr 25 '13 at 1:45
  • $\begingroup$ Yes, your edited answer is very helpful. Thanks. $\endgroup$
    – Gorg
    Apr 25 '13 at 19:48

@Gorg, you are on the right track. You can either solve this problem using small y or large Y. Your equations are set up correctly with small y, and the answer I get if you want to compare with what you get is $$x=135 \text{ and}\ y=180$$. Good job :)


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