1. Optimization Problems

1. A farmer has 3000yd of fence that he wishes to use to fence off an rectangular plot along the straight bank of a river. What are the dimensions of the maximum area he can enclose? What is this area?

2. If an open box has square base and a volume of 108 inches cubed and is constructed from a tin shieet, find the dimensions of the box, assuming a minimum amount of material is used in its construction.

3. By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting flaps, the cardboard may be turned into an open box. if the cardboard is 16inches wide, find the dimensions of the box that will yield maximum volume.

2. The 1st one

$p=3000=2l+w \iff w=3000-2l \mbox{ and } A=lw$

So if we sub w into the 2nd equaiton we get

$A=l(3000-2l)=2l(1500-l) -2l^2+3000l$

Since this is a parobola the vertex is halfway between the l-intercepts at 750
or we could use the vertex fomula

$l=\frac{-b}{2a}=\frac{-3000}{2(2)}=750$

Solving for w gives 1500.

3. #2

$V=108=x^2y \iff y=\frac{108}{x^2}\mbox{ and } S=x^2+4xy$

Using the above sub we get...

$S=x^2+x \cdot \frac{108}{x^2}=x^2+\frac{108}{x}$

Now taking the derivative we get

$\frac{dS}{dx}=2x-\frac{108}{x^2}$

setting equal to zero and solving we get

$0=2x-\frac{108}{x^2} \iff 0=2x^3-108 \iff 0=x^3-54$

so $x=3\sqrt[3]{2} \approx 3.78$

$y=\frac{108}{(3\sqrt[3]{2})^2}=\frac{12}{\sqrt[3]{4}} \approx 7.56$