# Optimization/Differentiation problem

• Dec 18th 2008, 05:20 AM
RealQuick
Optimization/Differentiation problem
1) A rancher wants to fence in an area of 3000000 square feet in a rectangular field and then divide it in half with a fence down the middle parallel to one side. What is the shortest length of fence that the rancher can use?

2) If 1100 square centimeters of material is available to make a box with a square base and an open top, find the largest possible volume of the box.

Can anyone help me with this?
• Dec 18th 2008, 10:57 AM
Greengoblin
1.) Let a and b be the sides of the rectangle:

$\displaystyle 3000000=ab \Rightarrow b=\frac{3000000}{a}$

The length of fence is given by:

$\displaystyle l=3a+2b=3a+\frac{6000000}{a}$

$\displaystyle 0=l'=3-\frac{6000000}{a^2}$

$\displaystyle a=\sqrt{2000000}=1414.2\text{ft} \Rightarrow b= \frac{6000000}{1414.2}=4242.6\text{ft}$

Therefore:

$\displaystyle l_{min}=3(1414.2)+2(4242.6)=12727.8\text{ft}$

2.) Let x,y,z, be the dimensions of the box and V the volume.

$\displaystyle 1100=xy+2xz+2yz$

Using the AM-GM inequality:

$\displaystyle \frac{xy+2xz+2yz}{3}\ge (xy\cdot 2xz\cdot 2yz)^{1/3}$

$\displaystyle \frac{1100}{3}\ge (xy\cdot 2xz\cdot 2yz)^{1/3}$

$\displaystyle \sqrt[3]{\frac{1100}{3}}\ge xy\cdot 2xz\cdot 2yz$

$\displaystyle \sqrt[3]{\frac{1100}{3}}\ge 4x^2y^2z^2$

$\displaystyle \sqrt[6]{\frac{1100}{3}}\ge 2xyz$

$\displaystyle V_{max}=xyz=\frac{\sqrt[6]{1100/3}}{2}=1.34\text{cm}^3$