Minimax problems.

For such problems, follow the flow chart:

\begin{center}Read the text twice
...rema of $f$
Draw a conclusion

Figure 15: The dimensions of a rectangle.

Example 6.11.1   Find the dimensions of a rectangle with maximal area and a perimeter of $ p$ meters.

Let $ x$ be the length of the rectangle; the width $ y$ verifies the relation: $ y= \frac 12 (p- 2x)$ . The area of the rectangle is given by: $ A(x)= xy= x \cdot \frac 12 (p- 2x)$ .

Then: $ A'(x)=\frac p2 -2x$ and $ A'(x)=0$ if, and only if, $ x= \frac p4$ .

As $ A(x)$ is a quadratic polynomial with negative main coefficient, the function $ A$ has a maximum at the point $ \frac p4$ . replacing in the definition of $ y$ , we have: $ x=y=\frac p4$ .

In order to have maximal area with a fixed perimeter, the rectangle should be a square.

Example 6.11.2   Consider a square plate whose side is 20 cm long. From each corner, cut a square in order to create a topless box by folding the sides upwards. Find the dimensions of the box for the volume to be maximal, then compute the maximal volume.

Denote by $ h$ the side length of the small squares cut off at the corners of the plate. This provides a box whose volume is given by the function $ \mathbb{S}$ defined as follows:

$\displaystyle \mathbb{S}(h)=(20-2h)^2*h = 4h^3-80h^2+400h.$    

Now we have:

$\displaystyle \mathbb{S}'(h)=12h^2-160h+400.$    


$\displaystyle \mathbb{S}'(h)=0 \Longleftrightarrow h=10$   or$\displaystyle \quad h=\frac{10}{3}.$    

If $ h=10$ , nothing is left and no box is built. Therefore $ h=10/3$ .

Apply the second derivative test:

$\displaystyle \mathbb{S}''(h)=24h-160$    

and $ \mathbb{S}''(10/3)=-80$ . As this number is negative, we have a confirmation that for $ h=10/3$ the volume of the box is maximal.

the maximal volume is given by:

$\displaystyle \mathbb{S}(10/3)=\frac {16000}{27}\approx 592.59 \;$   cm$\displaystyle ^3.$    

Noah Dana-Picard 2007-12-28