The volume of an open-topped rectangular box is 24 cm³,and the length of the rectangular base is twice as long as its width.What is the width of the base so that the surface area of the open box is minimized?

Question

Quizplus · Accepted Answer

Let the width of the base be $x$ cm.
Then the length of the base is $2x$ cm.
If the height of the box is $h$ cm, then we have:

$$24 = x\cdot 2x\cdot h = 2x^2h$$

Solving for $h$, we get $h=\frac{12}{x^2}$.
The surface area of the open box is given by:

$$A = 2xh+4x^2 = 2x\cdot \frac{12}{x^2}+4x^2 = \frac{24}{x}+4x^2$$

To minimize the surface area, we take the derivative with respect to $x$ and set it equal to zero:

$$-\frac{24}{x^2}+8x = 0$$

Solving for $x$, we get $x^3=3$, or $x=\sqrt[3]{3}$.
Therefore, the width of the base should be $\sqrt[3]{3}$ cm.

The Volume of an Open-Topped Rectangular Box Is 24 Cm3,and

The Volume of an Open-Topped Rectangular Box Is 24 Cm³,and