Step 1:
Create a drawing of the problem material.
Step 2:
The identify the variable and the interval which applies to the independent variable:Volume is equal to length times width times height (V = lwh).
V = V (x)
w = (8 2x)
l = (15 2x)
h = x
The interval of x is (0, 4).
Step 3:
Set up the equation, which models the data in the problem.V (x) = x(8 2x) (15 2x)
V(x) = 4x³ - 46x² + 120x
Step 4:
Graph the functions. V(x) = 4x³ - 46x² + 120x and y = 2450/27.Domain interval is [0, 10].
Domain interval is [0, 4].
Step 5:
First Derivative test for Maximum values.V(x) = 12x² - 92x + 120
V(x) = 4(3x² - 23x + 30)
V(x) = 4(3x 5)(x - 6)
x = 5/3 or 6
x
= 5/3The interval is [0, 4] and 6 is not an element of the interval, therefore 5/3 inches is the length of each cut that will maximize the volume for the box. The First Derivative Test and Closed Interval Test are two additional tests used to find optimal maximum and minimum value.
Step 6
: The maximum Volume is found by substituting 5/3 into the position function for each and every x.V(x) = 4x³ - 46x² + 120x
V(x) = x (15 2x) (8 2x)
V(5/3) = (15 2(5/3)) (8 2(5/3))
V(5/3) = (5/3 )(35/3) (14/3)
V(5/3) = 2450/27
Conclusion: The cutouts will be 5/3 inches by 5/3 inches. The dimensions of the box are 5/3 in. by 35/3 in. by 14/3 in. The maximum volume of this box is 2450/27inches³.