# Math 22 Optimization Problems

## Solving Optimization Sample Problems

1) Maximum Area: Find the length and width of a rectangle that has 80 meters perimeter and a maximum area.

Solution:
Let $l$ be the length of the rectangle in meter.
and $w$ be the width of the rectangle in meter.
Then, the perimeter $P=2(l+w)=80$ , so $l+w=40$ , then $l=40-w$ Area $A=l.w=(40-w)w=40w-w^{2}$ $A'=40-2w=0$ , then $w=20$ , so $l=40-w=40-20=20$ Therefore, $l=w=20$ 2) Maximum Volume A rectangular solid with a square base has a surface area of $337.5$ square centimeters. Find the dimensions that yield the maximum volume.

Solution:
Let $a$ be the length of the one side of the square base in centimeter.
and $h$ be the height of the solid in centimeter.
Then, the surface area $S=2a^{2}+4ah=337.5$ , so $h={\frac {337.5-2a^{2}}{4a}}$ Volume $V=a^{2}h=a^{2}({\frac {337.5-2a^{2}}{4a}})={\frac {1}{4}}a(337.5-a^{2})={\frac {337.5a}{4}}-{\frac {a^{3}}{4}}$ $V'={\frac {337.5}{4}}-{\frac {3}{4}}a^{2}=0$ , then $a^{2}={\frac {225}{4}}$ , so $a=\pm {\frac {25}{2}}={\frac {25}{2}}$ since $a$ is positive.
Hence, $h={\frac {337.5-2a^{2}}{4a}}=h={\frac {337.5-2({\frac {25}{2}})^{2}}{4{\frac {25}{2}}}}={\frac {1}{2}}$ Therefore, the dimensions that yield the maximum value is $a={\frac {25}{2}}$ and $h={\frac {1}{2}}$ 3) Minimum Dimensions: A campground owner plans to enclose a rectangular field adjacent to a river. The owner wants the field to contain $180000$ square meters. No fencing is required along the river. What dimensions will use the least amount of fencing?

Solution:
Let $a$ be the length of two sides that are connected to the river.
and $b$ be the length of the sides that is opposite the river.
Then, the area $A=ab=180000$ , so $b={\frac {180000}{a}}$ The fence $F=2a+b=2a+{\frac {180000}{a}}$ $F'=2-{\frac {18000}{a^{2}}}=0$ , then $a^{2}=9000$ , so $a={\sqrt {9000}}=\pm 30=30$ since $a$ is positive. Then, $b={\frac {180000}{30}}=6000$ Therefore, the dimensions of the fence is $a=30$ meters and $b=6000$ meters.