Answer :
To answer this question it is necessary to find the volume of the box as a function of "x", and apply the concepts of a maximum of a function.
The solution is:
a) V (max) = 36.6 in³
b) x = 1.3 in
The volume of a cube is:
V(c) = w×L×h ( in³)
In this case, cutting the length "x" from each side, means:
wide of the box ( w - 2×x ) equal to ( 7 - 2×x )
Length of the box ( L - 2×x ) equal to ( 9 - 2×x )
The height is x
Then the volume of the box, as a function of x is:
V(x) = ( 7 - 2×x ) × ( 9 -2×x ) × x
V(x) = ( 63 - 14×x - 18×x + 4×x²)×x
V(x) = 4×x³ - 32×x² + 63×x
Tacking derivatives, on both sides of the equation
V´(x) = 12×x² - 64 ×x + 63
If V´(x) = 0 then 12×x² - 64 ×x + 63 = 0
This expression is a second-degree equation, solving for x
x₁,₂ = [ 64 ± √ (64)² - 4×12*63
x₁ = ( 64 + 32.74 )/ 24
x₁ = 4.03 this value will bring us an unfeasible solution, since it is not possible to cut 2×4 in from a piece of paper of 7 in ( therefore we dismiss that value)
x₂ = ( 64 - 32.74)/24
x₂ = 1.30 in
The maximum volume of the box is:
V(max) = ( 7 - 2.60) × ( 9 - 2.60)×1.3
V(max) = 4.4 × 6.4 × 1.3
V(max) = 36.60 in³
To chek for maximum value of V when x = 1.3
we find the second derivative of V V´´, and substitute the value of x = 1.3, if the relation is smaller than 0, we have a maximum value of V
V´´(x) = 24×x - 64 for x = 1.3
V´´(x) = 24× 1.3 - 64 ⇒ V´´(x) < 0
Then the value x = 1.3 will bring maximum value for V
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