I need to maximize the volume of this box.?

Question

i start out with a 11 X 13 piece of carboard. equal squares are cut off each corner and the sides are turned up to form an open rectangular box.

i took calculus a while ago, and i pretty much forgot everything and i wanted to know how to go about this problem.

Answer 1

If the cut-out square has side length x, you'll have sides that are x tall when they are folded up. The lengths of the folded-up sides will be 11 - 2x and 13 - 2x. So the volume of the box will be V(x) = x(11 - 2x)(13 - 2x). Expand this expression to get a cubic equation.

The extreme values of a function always occur when the derivative is equal to zero, so find that derivative and set V'(x) = 0, and solve. The derivative of a cubic equation is a quadratic, so you should be able to find its zeroes using techniques you already know. You may get two different solution, and you'll need to check the values of V(x) at both of them to see which is larger; the other might represent a minimum or a local max, but not the absolute max. If one of the values for x is negative, you know immediately that it's not correct.

Answer 2

Dimensions of the box are 11 - 2x, 13 - 2x, and x (Draw the picture, you'll see). So, what you want is that x for which

V(x) = (11 - 2x)(13 - 2x)x

is maximum. For the problem to make sense, you need x between 0 and 11/2. If there is a maximum in this interval of x, it will be where the derivative V'(x) = 0 and the second derivative V''(x) < 0.

So, take the derivative of V(x), set it to zero and solve for x (you should have a quadratic equation). Pick that x which makes V''(x) negative, and that's your answer.

Answer 3

If x = side of the square you cut off
Vol, V = x (13-2x) (11-2x) = 143x - 48x^2 + 4x^3
dV/dx = 143 - 96x + 12x^2
dV/dx = 0, when x = 6.02, 1.98

x = 1.98 is the only realistic solution. This maximizes the volume, which is 126.01 cubic units

Answer 4

Let the width of the box be x,

The breadth as a function of x will be: y=2+x

And the height: z=(11-x)/2.


V=x(2+x)(11-x)/2

Find V for dV/dx=0 (and d²V/dx²<0)

Answer 5

v(x) = (11-2x)(13-2x)x, v(x) stands for the volume.
Solve v' = 0 for x,
x = 1.979
v(max) = v(1.979) = 126.0 units^3