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Can someone help me with this calculus question on optimisation?
An open box is made from a 4m^2 (4 meter squared) piece of cardboard by cutting small square pieces of side length x from each corner and folding up the sides.
a. Write down a formula for the volume V(x) on a suitable physical domain.
b. What is the maximum volume of the box? Justify.
Can someone please help with these 2 questions. thanks
1 Answer
- ?Lv 77 years ago
a.
The length of a side of the original square is 2 m, since the area is 4 m².
Since two pieces of length x are cut out of each side, the new length (in meters) is 2 - 2x.
When the sides are folded up, the base has an area of (2 - 2x)², or 4 - 8x + 4x².
The volume is therefore x times the base area: V = x(4 - 8x + 4x²) = 4x - 8x² + 4x³.
b.
The volume is maximized (or minimized) when its first derivative equals zero.
Take the derivative: V' = 4 - 16x + 12x².
Set it equal to zero:
12x² - 16x + 4 = 0, or
3x² - 4x + 1 = 0.
This factors as (3x-1)(x-1) = 0, so x = 1/3 or x = 1.
If x = 1, then the cutout part is the entire width of the square, since 2x = 2(1) = 2 is removed from the original length of 2. There would be nothing left to make the box, so ignore the x=1 root.
Instead, x = 1/3, and the volume is 4(1/3) - 8(1/3)² + 4(1/3)³
= 4/3 - 8/9 + 4/27
= 36/27 - 24/27 + 4/27
= 16/27 m³.
