An open box is made from a rectangular sheet of tin 16 cm long and 10 cm wide by cutting equal squares from the four corners and folding up the flaps. The maximum possible volume of the box is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 144 cm3
Let x cm be the side of each square cut off. The box has height x and base (16 − 2x) by (10 − 2x), so 0 < x < 5. Its volume is
\(V = x(16 - 2x)(10 - 2x)\)
\(V = 4x^3 - 52x^2 + 160x\)
Setting the derivative equal to zero:
\( \displaystyle \frac{dV}{dx} = 12x^2 - 104x + 160 \)
\( = 4(3x^2 - 26x + 40) \)
\( = 4(3x - 20)(x - 2) \)
So x = 2 or \(x = \frac{20}{3}\). The value \(\frac{20}{3}\) is more than 5, so it is not allowed (the width 10 − 2x would be negative). The second derivative test at x = 2:
\( \displaystyle \frac{d^2V}{dx^2} = 24x - 104 \)
\(\text{at } x = 2:\ 48 - 104 = -56 \lt 0\)
So V is a maximum at x = 2:
\(V = 2 \times 12 \times 6 = 144\) cm3