The answers for that question and probably many more are on this paper
6.875 inches, you divide the fraction to get a decimal
The questions for this problem would be:
1. What is the dimensions of the box that has the maximum volume?
2. What is the maximum volume of the box?
Volume of a rectangular box = length x width x height
From the problem statement,
length = 12 - 2x
width = 9 - 2x
height = x
where x is the height of the box or the side of the equal squares from each corner and turning up the sides
V = (12-2x) (9-2x) (x)
V = (12 - 2x) (9x - 2x^2)
V = 108x - 24x^2 -18x^2 + 4x^3
V = 4x^3 - 42x^2 + 108x
To maximize the volume, we differentiate the expression of the volume and equate it to zero.
V = 4x^3 - 42x^2 + 108x
dV/dx = 12x^2 - 84x + 108
12x^2 - 84x + 108 = 0x^2 - 7x + 9 = 0
Solving for x,
x1 = 5.30 ; Volume = -11.872 (cannot be negative)
x2 = 1.70 ; Volume = 81.872
So, the answers are as follows:
1. What is the dimensions of the box that has the maximum volume?
length = 12 - 2x = 8.60
width = 9 - 2x = 5.60
height = x = 1.70
2. What is the maximum volume of the box?
Volume = 81.872
As x increases by 1 the value of f(x) doubles so we need 2^x
When x = 0 the value of f(x) = 3
The correct answer is f(x) = 3(2^x). [the top option]
