3. Make a box (without a lid) of the greatest capacity from a piece of cardboard $30 \times 14$ (cm²) by cutting out equal squares from the corners and then folding up the sides of the cardboard to form the sides of the box (Fig. 70).
Solution. The area of the base of the box $S=(14-2 x)(30-2 x)$, the height $H=x$. The volume of the box $V'=S H=4 x(7-x)(15-x)$.

Let's investigate this function: $V'(x)=4\left(3 x^{2}-44 x+105\right)$. The roots of the derivative will be $x_{1}=11 \frac{2}{3}, x_{2}=3$.

Next, $V''(x)=8(3 x-22), V''(3)<0$. At the point $x=3$, we have a maximum. From the problem statement, it is clear that the value $x_{1}=11 \frac{2}{3}$ does not make sense for this problem.

Therefore, when $x=3$ cm, the maximum volume $V=576 \, \text{cm}^3$.

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-193.jpg?height=219&width=388&top_left_y=540&top_left_x=96)

Fig. 70

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-193.jpg?height=189&width=268&top_left_y=570&top_left_x=541)

Fig. 71

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-193.jpg?height=191&width=235&top_left_y=569&top_left_x=881)

Fig. 72