Answer:
The optimal volume is 63.1142 in3, with a cut of 1.5767 inches
Step-by-step explanation:
If we make a cut of x inches to create the box, the dimensions of the box will be:
Length = 10 - 2x
Width = 9 - 2x
Height = x
So the volume of the box would be:
Volume = (10 - 2x) * (9 - 2x) * x
Volume = 4x3 - 38x2 + 90x
To find the maximum volume, we need to take the derivative of the volume and find the values of x where it is equal to zero:
dV/dx = 12x2 - 76x + 90 = 0
6x2 - 38x + 45 = 0
Using Bhaskara's formula, we have:
Delta = 38^2 - 4*6*45 = 364
sqrt(Delta) = 19.08
x1 = (38 + 19.08) / 12 = 4.7567
x2 = (38 - 19.08) / 12 = 1.5767
Testing these values in the volume equation, we have:
Volume1 = 4*(4.7567)^3 - 38*(4.7567)^2 + 90*(4.7567) = -1.1883
(Negative value for the volume is not valid)
Volume1 = 4*(1.5767)^3 - 38*(1.5767)^2 + 90*(1.5767) = 63.1142
So the optimal volume is 63.1142 in3, with a cut of 1.5767 inches