Optimization Problems (College Board AP® Calculus BC): Revision Note

A large flower bed is being designed as a rectangle with a semicircle on each end, as shown in the diagram below.

The total area of the bed is to be$100\pi$ meters squared.

(a) Show that the perimeter of the bed is given by the formula

$P=\mathit{\pi }(r+\frac{100}{r})$

(b) Find the value of $r$ that minimizes the perimeter, and find the minimum value of the perimeter.

(c) Justify that this is the minimum perimeter.

Answer:

(a)

The width of the rectangle is $2r$ meters, and its length is $L$ meters

The area consists of a semi circle, plus a rectangle, plus another semi circle

$\frac{1}{2}\pi r^2+2rL+\hspace{0.17em}\frac{1}{2}\pi r^2=100\pi$

Simplify and write $L$ in terms of $r$

$\begin{array}{rcl}\pi r^2+2rL& =& 100\pi \\ 2rL& =& 100\pi -\mathit{\pi }{r}^2\\ L& =& \frac{50\pi }{r}-\frac{\pi }{2}r\end{array}$

The perimeter of the flower bed consists of two semi-circular arcs, and two straight lengths

$P=\pi r+\pi r+2L$

Substitute in the expression for $L$, and simplify to the desired expression for $P$

$\begin{array}{rcl}P& =& \pi r+\pi r+2\left(\frac{50\pi }{r}-\frac{\pi }{2}r\right)\\ P& =& 2\pi r+\frac{100\pi }{\mathrm{r}}-\pi r\\ P& =& \pi r+\frac{100\pi }{r}\end{array}$

$P=\mathit{\pi }(r+\frac{100}{r})$

(b)

To find the minimum value of $P$, we need to find the minimum point on a graph of $P$ against $r$

To do this, we need to differentiate $P$ with respect to $r$

Start by writing $P$ in terms of $r$, using the answer from part (a)

$\begin{array}{rcl}P& =& \pi (r+\frac{100}{r})\\ P& =& \pi r+100\pi r^{-1}\end{array}$

Differentiate $P$ with respect to $r$

$\frac{dP}{dr}=\pi -100\mathit{\pi }{r}^{-2}$

At the minimum point, the derivative will be zero (assuming the minimum is at a critical point)

$\begin{array}{rcl}\pi -100\mathit{\pi }{r}^{-2}& =& 0\\ \pi -\frac{100\pi }{r^2}& =& 0\\ 1-\frac{100}{r^2}& =& 0\\ 1& =& \frac{100}{r^2}\\ r^2& =& 100\\ r& =& \pm 10\end{array}$

$r$ is a length, so -10 can be ignored

$r=10$

Find the minimum value of the perimeter by substituting $r=10$ into the equation for $P$

$P=\mathit{\pi }\left(10+\frac{100}{10}\right)=20\pi$

$r=10$ meters minimizes the perimeter

The minimum value of the perimeter will be 20π meters

(c)

To prove it is a minimum, show that the second derivative is positive (and the graph is therefore concave up) at this point

Start by finding the second derivative

$$\begin{gathered} \frac{dP}{dr}=\pi -100\mathit{\pi }{r}^{-2} \\ \frac{d^{2}P}{d{r}^2}=200\mathit{\pi }{r}^{-3} \end{gathered}$$

Substitute in $r=10$

${\overline{)\frac{d^{2}P}{d{r}^2}}}_{r=10}=200\pi {\left(10\right)}^{-3}=\frac{200\pi }{1000}=\frac{\pi }{5}>0$

Note that we don't need to check endpoints here: as $r\to 0$ the perimeter function becomes unbounded, while as $r\to \infty$ the perimeter also goes to infinity

Second derivative is positive at $r=10$, therefore 20π  meters is the minimum value for the perimeter

A manufacturer produces custom electronic components. The daily cost to produce x batches of components, measured in dollars, is modeled by the function $C\left(x\right)=2x^3-21x^2+60x+100$ for the interval $1\le x\le 6$.

Find the absolute minimum and absolute maximum daily cost to produce the components on the closed interval $1\le x\le 6$. Justify your answers.

Answer:

Find the derivative of the function

$C\prime \left(x\right)=6x^2-42x+60$

Set the derivative equal to zero to find critical points and solve

$$\begin{gathered} C\text{'}\left(x\right)=0 \\ 6x^2-42x+60=0 \\ {x}^2-7x+10=0 \\ \left(x-2\right)\left(x-5\right)=0 \\ x=2,x=5 \end{gathered}$$

Evaluate the function at the critical points and the endpoints

$$\begin{gathered} C\left(1\right)=2{\left(1\right)}^3-21{\left(1\right)}^2+60\left(1\right)+100=2-21+60+100=141 \\ C\left(2\right)=2{\left(2\right)}^3-21{\left(2\right)}^2+60\left(2\right)+100=16-84+120+100=152 \\ C\left(5\right)=2{\left(5\right)}^3-21{\left(5\right)}^2+60\left(5\right)+100=250-525+300+100=125 \\ C\left(6\right)=2{\left(6\right)}^3-21{\left(6\right)}^2+60\left(6\right)+100=432-756+360+100=136 \end{gathered}$$

Justify the use of the candidates test

$C\left(x\right)$ is a polynomial, so it is continuous over $1\le x\le 6$

The extreme value theorem guarantees that $C$ has a global maximum and a global minimum

These occur at the critical points or at the endpoints

State the minimum and maximum values

The absolute minimum daily cost is $125 (when 5 batches are produced)

The absolute maximum daily cost is $152 (when 2 batches are produced)