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
The answer i came up with is : (4b-7x) (a+2)
Answer:
0.30
Step-by-step explanation:
You only need to significant figures which means two numbers following the decimal point for this problem. No rounding is necessary as no numbers are over five following your first two digits past the decimal place.
Shortest is 3/4 longest is 1 1/2. 1 1/2 -3/4=3/4
Answer:
m=-7/9
Step-by-step explanation:
