# 12 - 14 Rectangular Lot Problems in Maxima and Minima

**Problem 12**

A rectangular field of fixed area is to be enclosed and divided into three lots by parallels to one of the sides. What should be the relative dimensions of the field to make the amount of fencing minimum?

**Solution 12**

$A = xy$

$0 = xy' + y$

$y' = -y/x$

Fence:

$P = 2x + 4y$

$dP/dx = 2 + 4y' = 0$

$2 + 4(-y/x) = 0$

$y = \frac{1}{2}x$

width = ½ × length *answer*

**Problem 13**

Do Ex. 12 with the words "three lots" replaced by "five lots".

**Solution 13**

$A = xy$

$0 = xy' + y$

$y' = -y/x$

Fence:

$P = 2x + 6y$

$dP/dx = 2 + 6y' = 0$

$2 + 6(-y/x) = 0$

$y = \frac{1}{3} x$

$\text{width } = \frac{1}{3} \, \times \, \text{ length } \,\,$ *answer*

**Problem 14**

A rectangular lot is bounded at the back by a river. No fence is needed along the river and there is to be 24-ft opening in front. If the fence along the front costs \$1.50 per foot, along the sides \$1 per foot, find the dimensions of the largest lot which can be thus fenced in for \$300.

**Solution 14**

$300 = 2y + 1.5(x - 24)$

$y = 168 - 0.75x$

Area:

$A = xy$

$A = x (168 - 0.75x)$

$A = 168x - 0.75x^2$

$\dfrac{dA}{dx} = 168 - 1.5x = 0$

$x = 112 \, \text{ft}$

$y = 168 - 0.75(112)$

$y = 84 \, \text{ft}$

**Dimensions:** 84 ft × 112 ft *answer*