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
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
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.
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.