Answer:
x = 2.354 in
Maximum volume = 228.162 in³
Step-by-step explanation:
Size of the metal sheet = (18 in × 12 in)
Let the height of the box = x in
Then length of the box = (18 - 2x) in
Width of the box = (12 - 2x) in
Now volume of the rectangular box (V) = Length × Width × height
= (18 - 2x) × (12 - 2x) × x
= x[18(12 - 2x) - 2x(12 - 2x)]
= x[216 - 36x - 24x + 4x²]
= x[216 - 60x + 4x²]
= (4x³ - 60x² + 216x) in³
For maximum volume, we will find the derivative of the volume and equate it to zero.
[tex]\frac{dV}{dx}=\frac{d}{dx}(4x^3-60x^2+216x)[/tex]
V' = 12x² - 120x + 216
V' = 0
12x² - 120x + 216 = 0
x² - 10x + 18 = 0
By quadratic formula,
x = [tex]\frac{10\pm\sqrt{(-10)^2-4(1)(18)}}{2(1)}[/tex]
x = 5 ± √7
x = 7.646, 2.354
But for 7.646, volume of the box will be negative.
Therefore, (x = 2.354 in) will be value of x for maximum volume of the box.
Maximum volume of the box = 4(2.354)³ - 60(2.354)² + 216(2.354)
= 228.162 in³
