SOLUTION: I have a two part question. A box without a top is constructed from a 4m by 8m piece of cardboard by cutting a square of side length x from each corner. a. Find the volume (v)

Question 320895: I have a two part question.
A box without a top is constructed from a 4m by 8m piece of cardboard by cutting a square of side length x from each corner.
a. Find the volume (v) of this box as a function of x.
b. By sketching the graph of the function v(x), establish that v(x) attains a maximum value on the interval (0,2).
Answer by toidayma(44) (Show Source): You can put this solution on YOUR website!
after cutting for squares of side length x each corner, we have a box with height of x and base of (4-2x)(8-2x)
Of course, 0 <= x <= 2.
a) So, the Volume of the box is V(x) = x(4-2x)(8-2x)
b) the point is that to find the maximum of V(x) with x belonging to (0,2).
V(x) = x(4-2x)(8-2x) = 4(x^3 - 4x^2 - 2x + 8)
V'(x) = 4(3x^2 - 8x - 2), V'(x) = 0 <-> x1 = (8+sqrt(44/3))/6 or x2 = (8-sqrt(44/3))/6.
I hope that you can easily get here because it is just about solving the quadratic equation.
OK, by sketching the graph of V(x), you will easily see that V(x): increasing from x =0 to x = x2, then decreasing from x = x2 to x = 2. Briefly, V(x) get the maximum value at x =x2. (This value of x2 also belongs to (0,2), but x1 is larger than 2.)
I hope that this can guide you to the final answer of this problem. It is just one popular problem :-)
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