Maximize Volume of Open Box from Rectangular/Square Cardboard

Question

a) From a 16cm long, 10cm wide cardboard, cut squares of side x from each corner, fold up the sides to make an open box. For which x is the volume maximum? What is the maximum volume? b) Fold an optimal box from A4 paper; c) Find x for a square cardboard of side a.

Accepted Answer

a) $x≈1.81$ cm, maximum volume ≈ 146.06 cm³; c) $x=\frac{a}{6}$

Answer

Step 1 (part a): Define box dimensions: length = 16-2x, width=10-2x, height=x. Volume function $V(x)=x(16-2x)(10-2x)=4x^3-52x^2+160x$ Step 2 (part a): Find first derivative $V'(x)=12x^2-104x+160$, set to 0: $12x^2-104x+160=0$ → divide by 4: $3x^2-26x+40=0$ Step 3 (part a): Solve quadratic: $x=\frac{26\pm\sqrt{676-480}}{6}=\frac{26\pm\sqrt{196}}{6}=\frac{26\pm14}{6}$. Valid solution: $x=\frac{12}{6}=2$ (wait, correct calculation: $\sqrt{676-480}=\sqrt{196}=14$, so $x=\frac{26-14}{6}=2$, $x=\frac{26+14}{6}=\frac{40}{6}≈6.67$ (invalid, since 10-2x would be negative). Recalculate derivative: $V(x)=x(16-2x)(10-2x)=x(160-52x+4x²)=4x³-52x²+160x$, $V'(x)=12x²-104x+160$. Using quadratic formula, $x=\frac{104\pm\sqrt{10816-7680}}{24}=\frac{104\pm\sqrt{3136}}{24}=\frac{104\pm56}{24}$. $x=\frac{48}{24}=2$, $x=\frac{160}{24}≈6.67$. So x=2, volume=2*(12)*(6)=144 cm³; or using second derivative $V''(2)=24-104=-80<0$ (maximum) Step 1 (part c): For square cardboard of side a, volume $V(x)=x(a-2x)^2=x(a²-4ax+4x²)=4x³-4ax²+a²x$. Find derivative $V'(x)=12x²-8ax+a²$, set to 0: $12x²-8ax+a²=0$. Solve: $x=\frac{8a\pm\sqrt{64a²-48a²}}{24}=\frac{8a\pm\sqrt{16a²}}{24}=\frac{8a\pm4a}{24}$. Valid solution $x=\frac{4a}{24}=\frac{a}{6}$