Math 22 Optimization Problems

Solving Optimization 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 $h={\frac {1}{2}}$

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