An open box is to be made from a six inch by six inch piece of material by cutting equal squares from the corners and turning up
2 answers:
Answer: 16 in^3
Step-by-step explanation: So this is a fun little calculus optimization problem. First, we must make an equation describing the volume of the box. Each side is 6 inches starting and then we remove "x" from each side twice. Both sides can then be described as 6 - 2x. Then the height of the box will be "x" inches. Our total equation is then V = (6-2x)^2 • x. Lets FOIL this equation out.
(6-2x)(6-2x)•x = (4x^2 - 24x + 36) • x
Then multiply the x
V = 4x^3 - 24x^2 + 36x
So this equation describes the volume. We want to find the maximum volume possible. This will be at the highest point of the graph where the slope = 0. So we take the derivative.
dy/dx 4x^3 - 24x^2 + 36x = 12x^2 - 48x + 36.
Now we plug this into the quadratic formula to find the points where the slope (x) equals zero.
x = (48 +- )/24
sqrt(2304 - 1728) = sqrt(576) = 24
So x = (48 +- 24)/24
Our two answers are (48-24)/24 = 1 = x
and (48+24)/24 = 3 = x
These two answers will be our minimum and maximum values for volume.
Now we can think, if we take 3 inches from the sides of the box and we have two sides, there will be no volume, so that can't be correct. And we can test this by plugging 3 into the equation. 4(3)^3 - 24(3)^2 + 36(3) = 0.
So 1 must be the maximum. When we plug this into the equation we get
4(1)^3 - 24(1)^2 + 36(1) = 16.
So the answer is 16in^3 by taking 1 inch squares out. This makes each side 4 inches and the height 1 giving us the volume equation. 4in•4in•1in = 16 in^3.
Please comment if anything needs clarified.
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