# maxima and minima

• Feb 24th 2008, 11:33 PM
icyxbluxkitty
maxima and minima
i really need help for the following problems in finding their maxima and minima

1. a rectangular box with a square box has the top and bottom made of one material while the sides are made of a different material which costs twice as much. the volume is 100 cubic cm. find the dimensions so that the cost of production would be as low as possible.

2. Leo is at pt. A walking eat at 2m/s. Aya who is 60 m North East of Leo starts walking North at the same instant at the rate of 3 m/s. how fast are they separating one minute later?

3. f(x) = 2 sin(3x)

and i know this sounds dumb, but how would u solve

(-x^2 + 2x + 2 ) / (x^2 + 2) ^2 = 0

thanks very much if you can help...>.< im going crazy over this. if anyone can solve any of the problems please...even if just one is solved..:)
• Feb 25th 2008, 03:42 AM
CaptainBlack
Quote:

Originally Posted by icyxbluxkitty
and i know this sounds dumb, but how would u solve

(-x^2 + 2x + 2 ) / (x^2 + 2) ^2 = 0

You can multiply through by $(x^2+2)^2$ to get:

$-x^2 + 2x + 2 = 0$

RonL
• Feb 25th 2008, 11:28 AM
earboth
Quote:

Originally Posted by icyxbluxkitty
...

1. a rectangular box with a square box (do you mean bottom) has the top and bottom made of one material while the sides are made of a different material which costs twice as much. the volume is 100 cubic cm. find the dimensions so that the cost of production would be as low as possible.

...

I assume that the box has a bottom like a square.

Let x denote the length of one side of the square and h denote the height of the box.

Then you know:
A) volume:
$V = x^2 \cdot h = 100$
B) surface area:
$s = 2x^2+4x \cdot h$
C) costs:
$c = 2x^2 + 2 \cdot 4x \cdot h$

From A) you get: $h=\frac{100}{x^2}$ . Plug in this term instead of h into the equation at C):

$c(x) = 2x^2 + \frac{800}{x}$

The function c(x) has an extreme value (minimum or maximum) if c'(x) = 0:

$c'(x) = 4x - \frac{800}{x^2}$ . Set c'(x) = 0 and solve for x:

I've got $x = \sqrt[3]{200} \approx 5.848$

Now plug in this value into the equation calculating h:

$h=\frac{100}{(\sqrt[3]{200})^2} = \sqrt[3]{\frac{1,000,000}{40,000}} = \sqrt[3]{25} = \sqrt[3]{\frac18 \cdot 200} = \frac12 \cdot \sqrt[3]{200}$ and therefore $h = \frac12 x$