# More Optimization Problems(plz Help!)

• Mar 27th 2008, 09:33 PM
sweetchocolat113
More Optimization Problems(plz Help!)
1. Find the minimum value of A if A=4y + x^2, where (x^2 + 1)y=324.

2. 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 16 inches long and 10 inches wide, find the dimensions of the box that will yield maximum volume
• Mar 28th 2008, 08:44 AM
iknowone
1. Since you are given the restriction that (x^2+1)y=324 solve for y and substitute into the formula for A.

y = 324/(1+x^2)

A
= 4(324/(1+x^2)) + x^2
= 1296/(1+x^2) + x^2

Differentiate and solve x in the equation A'(x)=0. This will give you the local maximum or minimum values of A(x). Decide which is a max and which is a min if necessary to find the correct solution. Use this value of x to solve for y and A.

2. So the rectangle has dimensions l and w. Say the squares have side length x. Then the box will have the following dimensions:

height: x
length: l-2x
width: w-2x
volume: x(l-2x)(w-2x)

Differentiate volume and solve for x in the equation V'(x) = 0.
• Mar 28th 2008, 01:35 PM
Soroban
Hello, sweetchocolat113!

A good sketch will help . . .

Quote:

2. 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 16 inches long and 10 inches wide,
find the dimensions of the box that will yield maximum volume.

Code:

      : - - -  16 - - - - :     - *---*-----------*---* -     : |///:          :///| x     : | - + - - - - - + - | -     : |  :          :  | :   10 |  :          :  | 10-2x     : |  :          :  | :     : | - + - - - - - + - | -     : |///:          :///| x     - *---*-----------*---* -       : x : - 16-2x - : x :
The cardboard is 10 by 16 inches.
$x$-by- $x$ squares are cut from each corner.
The "flaps" are folded up to form an open-top box.

Code:

        *-----------*         /|          /|       / |        / |x       *-----------*  |       |          |  *     x|          | /10-2x       |          |/       * - - - - - *         16-2x

The volume of the box is: . $V \;=\;L\!\cdot\!W\!\cdot\!H \;=\;(16-2x)(10-2x)x$

Therefore, we must maximize the function: . $V \;=\;160x - 52x^2 + 4x^3$

Go for it!