Answer:
The length of the box is 2×√30 inches long, the width of the box is 2×√(10/3) inches long, while the height of the box is √30 inches long
Explanation:
The given parameters are;
The length of the last rectangular prism box = 3 × The width of the box
The area of the available wrapping paper = 240 in.²
Let L represent the length of the rectangular prism box, let W represent the width of the rectangular prism box and let h, represent the height of the rectangular box
Therefore, we have;
L = 3 × W
The surface area of the box, A = 2 × L × W + 2 × L × h + 2 × W × h
Whereby, L = 3 × W, we have;
A = 2 × 3 × W × W + 2 × 3 × W × h + 2 × W × h = 8·h·W + 6·W²
A = 8·h·W + 6·W² = 240
8·h·W + 6·W² = 240
8·h·W = 240 - 6·W²
h = (240 - 6·W²)/(8·W)
The volume, V = L × W × h
∴ V = 3 × W × W × h = 3 × W² × h
V = 3 × W² × h
∴ V = 3 × W² × (240 - 6·W²)/(8·W) = 3/8 × W × (240 - 6·W²) = 90·W - 9/4·W³
dV/dW = d(90·W - 9/4·W³)/dW = 90 - 27/4·W²
At the maximum point, dV/dW = 0, which gives;
dV/dW = 90 - 27/4·W² = 0 at maximum point
27/4·W² = 90
W² = 4/27 × 90 = 40/3
W = ±√(40/3) = ±2×√(10/3)
Differentiating again and substituting gives;
d²V/dW² = d(90 - 27/4·W²)/dW = - 27/4·W
Given that when W = 2×√(10/3), d²V/dW² is negative, and when W = -2×√(10/3), d²V/dW² is positive, W = 2×√(10/3) is the local maximum point
Therefor, the dimensions of the box are;
W = 2×√(10/3)
L = 3 × W = 3 × 2×√(10/3)
L = 3 × 2×√(10/3) = 2×√30
L = 2×√30
L = 6 ×√(10/3)
h = (240 - 6·(2×√(10/3))²)/(8·(2×√(10/3))) = √30
h = √30
Therefore, the length of the box is 2×√30 inches long, the width of the box is 2×√(10/3) inches long, while the height of the box is √30 inches long.