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:
$$ 4x+3y = 1080,~~~\text{Area} = 3xy.$$
Best Answer
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 $180 \text{ ft} \times 135\text{ ft}$