Calculus one optimization problem HELP
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Calculus one optimization problem HELP

[From: ] [author: ] [Date: 12-07-17] [Hit: ]
(what???) find the largest volume that such a box can have. let x denote the length of the side of the square being cut out. let y denote the length of the base.......
im just not sure how to set up my constraint and my objective equations because the problem is really hard for me to visualize and understand.

a box with an open top is to be constructed from a square piece of cardboard, 3 feet wide, by cutting out a square from each of the four corners and bending up the sides. (what???) find the largest volume that such a box can have. let x denote the length of the side of the square being cut out. let y denote the length of the base.

write an expression for the volume:
V = ?
use the given info to write an equation that relates the variables and then write volume as a function of x

V(x) =

then find the largest volume

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If you are having trouble visualizing taking a square of cardboard and making an open-topped box from it, then go in reverse.

Think of an open top box. Make it a square box, while you are at it. Now, cut the sides straight down at the four corners. When it falls flat, you have a big cross-shaped piece of cardboard, which is just missing some cutouts from each corner of the original cardboard sheet. I hope that helps?

If squares are being cut from each corner, and each square has a side of length x, then, x will become the height of the box.

y will equal the sheet length, 3 feet, minus 2x. Since every part of this problem is a square, then, y will be the width and the depth of the open box.

V = y * y * x
y = 3 ft - 2x

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V(x) = x(3-2x)^2 = 9x -12x^2 + 4x^3
To find the largest volume, differentiate:
V'(x) = 9 -24x + 12x^2 = 0
4x^2 -8x + 3 = 0
x = 1/2, x = 3/2
So, the solution is x = 1/2 ( the solution x = 3/2 should be ruled out- you figure out why!)
and the max. volume V = 1 ft^2.
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