A square sheet of cardboard 48 inches on a side is made into a box by cutting squares of equal size from each corner of the sheet and folding the projecting flaps into an open-top box. What should be the length of the edge of any of the cutout squares to give the box maximum volume?
2006-10-22 04:21:39 · 5 answers · asked by chris 2 in Science & Mathematics ➔ Mathematics
16 inches
2006-10-22 04:30:24 · answer #1 · answered by CF_ 7 · 0⤊ 1⤋
Let x be the length of the edge of any of the cutout squares.
Then the base of the box will have length and width (48-2x), and the height will be x. Therefore the volume will be x(48-2x)^2. Certainly x must lie in the interval (0,24).
To find the maximum volume, differentiate and set the derivative equal to zero. Using the product rule, you get the equation
(48-2x)^2 + 2x(48-2x)(-2) = 0.
Simplify the left side:
(48-2x)(48-2x-4x) = 0
becomes
(48-2x)(48-6x) = 0
so the derivative is zero when 48-2x=0 or 48-6x=0. That is, when x=24 or x=8. x=24 makes no sense, because you'd have nothing left. Therefore x=8 is the only candidate for a local maximum. This is also the absolute maximum on the interval [0,24], because the volume is 0 when x=0 or x=24.
2006-10-22 04:40:16 · answer #2 · answered by James L 5 · 1⤊ 0⤋
Let x = the size of the the square being cut off.
Then the volume of the box will be x(48-2x)^2 =
48^2 x - 192x^2 +4 x^3
So dV/dx = 48^2 -384x +12x^2
To find max, set the above to zero and solve for x
x^2 - 32x + 192 = 0
(x-24)(x-8)=0
So x = 8 or x =24
But x= 24 is a minimum which leaves x= 8 inches as the correct answer.
2006-10-22 05:24:22 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
If you cut out 16" squares, you get a 16 x 16 x 16 box. 4096 cu" If you cut out 8" squares, you get a 32 x 32 x 8 " box. 8192 cu"
So much for math democracy.
2006-10-22 05:34:00 · answer #4 · answered by Nomadd 7 · 0⤊ 0⤋
16 in.
2006-10-22 04:31:26 · answer #5 · answered by Candy!!!!!!!!! 4 · 0⤊ 1⤋