# 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 ${\displaystyle l}$ be the length of the rectangle in meter.
and ${\displaystyle w}$ be the width of the rectangle in meter.
Then, the perimeter ${\displaystyle P=2(l+w)=80}$, so ${\displaystyle l+w=40}$, then ${\displaystyle l=40-w}$
Area ${\displaystyle A=l.w=(40-w)w=40w-w^{2}}$
${\displaystyle A'=40-2w=0}$, then ${\displaystyle w=20}$, so ${\displaystyle l=40-w=40-20=20}$
Therefore, ${\displaystyle l=w=20}$

2) Maximum Volume A rectangular solid with a square base has a surface area of ${\displaystyle 337.5}$ square centimeters. Find the dimensions that yield the maximum volume.

Solution:
Let ${\displaystyle a}$ be the length of the one side of the square base in centimeter.
and ${\displaystyle h}$ be the height of the solid in centimeter.
Then, the surface area ${\displaystyle S=2a^{2}+4ah=337.5}$, so ${\displaystyle h={\frac {337.5-2a^{2}}{4a}}}$
Volume ${\displaystyle V=a^{2}h=a^{2}({\frac {337.5-2a^{2}}{4a}})={\frac {1}{4}}a(337.5-a^{2})={\frac {337.5a}{4}}-{\frac {a^{3}}{4}}}$
${\displaystyle V'={\frac {337.5}{4}}-{\frac {3}{4}}a^{2}=0}$, then ${\displaystyle a^{2}={\frac {225}{4}}}$, so ${\displaystyle a=\pm {\frac {25}{2}}={\frac {25}{2}}}$ since ${\displaystyle a}$ is positive.
Hence, ${\displaystyle h={\frac {337.5-2a^{2}}{4a}}=h={\frac {337.5-2({\frac {25}{2}})^{2}}{4{\frac {25}{2}}}}={\frac {1}{2}}}$
Therefore, the dimensions that yield the maximum value is ${\displaystyle a={\frac {25}{2}}}$ and ${\displaystyle h={\frac {1}{2}}}$

3) Minimum Dimensions: A campground owner plans to enclose a rectangular field adjacent to a river. The owner wants the field to contain ${\displaystyle 180000}$ square meters. No fencing is required along the river. What dimensions will use the least amount of fencing?

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