V = (12 – 2x)(8 – 2x)(x)
V = 96x – 40x^2 + 4x^3
To find the zeros, we equate the equation to 0, so, the values of x that would result to zero would be:
x = 0, 6, 4
2. To get the value of x to obtain the maximum capacity, we differentiate the equation, equate it to zero, and solve for x.
dV/dx = 96 – 80x + 12x^2 = 0
x = 5.10, 1.57
The value of x that would give the maximum capacity is x = 1.57
3. If the volume of the box is 12, then the value of x can be solved using:
12 = 96x – 40x^2 + 4x^3
x = 0.13, 6.22, 3.65
The permissible value of x is 0.13 and 3.65
4. Increasing the cutout of the box increases the volume until its dimension reaches 1.57. After that, the value of the volume decreases it reaches 4.
5. V = (q -2x) (p – 2x) (x)