Problem 20. From a square sheet of metal with a width of 60 cm, a box without a lid and with a square base of the largest possible volume needs to be made.

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\text { Solution. }
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Let the width of the square strips to be folded up be $x$. Note that $x<\frac{60}{2}=30$. The width of the square base is $60-2 x$. We find the volume of the box:

$$
V=(60-2 x)^{2} x
$$

Multiplying both sides of equation (1) by 4, we get:

$$
4 V=(60-2 x)(60-2 x) \cdot 4 x
$$

The sum of the (positive) factors on the right side of equation (2) is 120 (it is significant that two of the three factors are the same). Therefore, the product of these factors reaches its maximum value when $60-2 x=4 x$, i.e., $x=10$.

It is precisely then that $4 V$, and thus $V$, reach their maximum values:

$$
V_{\text {max }}=40 \cdot 40 \cdot 10=16000\left(\mathrm{~cm}^{3}\right) .
$$