# 09 - 11 Rectangular Lot Problems in Maxima and Minima

**Problem 9**

What should be the shape of a rectangular field of a given area, if it is to be enclosed by the least amount of fencing?

**Solution 9**

$A = xy$

$0 = xy' + y$

$y' = -y/x$

Perimeter:

$P = 2x + 2y$

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

$1 + (-y/x) = 0$

$y = x$ (a square) *answer*

**Problem 10**

A rectangular field of given area is to be fenced off along the bank of a river. If no fence is needed along the river, what is the shape of the rectangle requiring the least amount of fencing?

**Solution 10**

$A = xy$

$0 = xy' + y$

$y' = -y/x$

Perimeter:

$P = x + 2y$

$dP/dx = 1 + 2y' = 0$

$1 + 2(-y/x) = 0$

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

width = ½ × length *answer*

**Problem 11**

A rectangular lot is to be fenced off along a highway. If the fence on the highway costs m dollars per yard, on the other sides n dollars per yard, find the area of the largest lot that can be fenced off for k dollars.

**Solution 11**

$k = mx + n(2y + x)$

$k = mx + 2ny + nx$

$k - (m + n)x = 2ny$

$y = \dfrac{k}{2n} - \dfrac{m + n}{2n}x$

Area:

$A =xy$

$A = x\left( \dfrac{k}{2n} - \dfrac{m + n}{2n}x \right)$

$A = \dfrac{k}{2n}x - \dfrac{m + n}{2n}x^2$

$\dfrac{dA}{dx} = \dfrac{k}{2n} - \dfrac{m + n}{n}x = 0$

$\dfrac{k}{2n} = \dfrac{m + n}{n}x$

$x = \dfrac{k}{2(m + n)}$

$y = \dfrac{k}{2n} - \dfrac{m + n}{2n} \, \left[ \dfrac{k}{2(m + n)} \right]$

$y = \dfrac{k}{4n}$

$A = \dfrac{k}{2(m + n)} \times \dfrac{k}{4n}$

$A = \dfrac{k^2}{8n\,(m + n)} \, \text{ yard}^2$ *answer*

- Facebook Like
- Log in or register to post comments
- Email this page
- 49908 reads