MATH 221

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updated: 14 August 2014

Lecture 20

Optimization

Critical points are where maxima and minima might occur.

Find the local maxima and minima of $f\left(x\right)=2x^3-24x+107$ in the interval $[1,3]\text{.}$

The critical points are

(a)	points where $\frac{df}{dx}$ is $0,$ $$\begin{array}{c} \end{array}$$
(b)	points where $f\left(x\right)$ is not continuous or not differentiable, $$\begin{array}{c}\end{array}$$
(c)	points on the boundary of where $f\left(x\right)$ is defined $$\begin{array}{c} \end{array}\text{or}\begin{array}{c} \end{array}\text{or}\begin{array}{c} \end{array}\text{or...}$$

For $f\left(x\right)=2x^3-24x+107$ in the interval $[1,3],$ $x=1$ and $x=3$ are critical points of type (c), and $\frac{df}{dx}=6x^2-24$ and $6x^2-24=0$ when $x^2=\frac{24}{6}=4\text{.}$ So $x=\pm 2$ is when $\frac{df}{dx}$ is $0\text{.}$ So $x=2$ is a critical point in $[1,3]\text{.}$

Critical point $x=1\text{:}$ $$\frac{df}{dx}{|}_{x=1}=6x^2-24{|}_{x=1}=6-24<0\text{.}$$ So $f\left(x\right)$ is decreasing at $x=1\text{.}$ So (from the picture) $x=1$ is a maximum. $$\begin{array}{c} 1 3 x y \end{array}$$ Critical point $x=3\text{:}$ $$\frac{df}{dx}{|}_{x=3}=6x^2-24{|}_{x=3}=6·3^2-24=30>0\text{.}$$ So $f\left(x\right)$ is increasing at $x=3\text{.}$ So (from the picture) $x=3$ is a maximum. $$\begin{array}{c} 1 3 x y \end{array}$$ Critical point $x=2\text{:}$ $$\frac{df}{dx}{|}_{x=2}=0,\frac{d^{2}f}{d{x}^2}{|}_{x=2}=12x{|}_{x=2}=24>0\text{.}$$ So $f\left(x\right)$ is flat and concave up at $x=2\text{.}$ So $x=2$ is a minimum. $$\begin{array}{c} 1 2 3 x y \end{array}$$

An enemy jet is flying along the curve $y=x^2+2\text{.}$ A soldier is placed at the point $(3,2)\text{.}$ At what point will the jet be at when the soldier and the jet will be the closest? $$\begin{array}{c} -1 1 2 3 x 1 2 3 y d (3,2) \end{array}$$ If the jet is at the point $(p,q)$ the distance between them is $$d=\sqrt{{(p-3)}^2+{(q-2)}^2}\text{.}$$ The point $(p,q)$ is on the curve $y=x^2+2$ so $q=p^2+2\text{.}$ So $d=\sqrt{{(p-3)}^2+{(p^2+2-2)}^2}\text{.}$ We want to minimize $d$ (as the jet moves, i.e. as $p$ changes). The distance $d$ will be minimum at the same time that $d^2$ will be minimum. So we can minimize $d^2\text{.}$ $$d^2={(p-3)}^2+{\left(p^2\right)}^2={(p-3)}^2+p^4\text{.}$$ Find a critical point. When is $$\frac{d{d}^2}{dp}=2(p-3)+4p^3=4p^3+2p-6=(p-1)(4p^2+4p+6)$$ equals to $0\text{??}$ When $p=1\text{.}$ So $\frac{d{d}^2}{dp}{|}_{p=1}=0,$ so $p=1$ is a critical point. From the picture ew can confirm that when the jet is at $(1,3)$ (i.e. $p=1,$ $q=3\text{)}$ the distance to the soldier is minimum.

Maximize the volume of a cone with a given slant height. Show that the angle of inclination is ${\text{tan}}^{-1}\sqrt{2}\text{.}$ $$\begin{array}{cc}\begin{array}{c} ℓ ℓ h θ r \end{array}& \begin{array}{ccc}{\displaystyle \ell }& =& {\displaystyle \text{slant height,}}\\ {\displaystyle \theta }& =& {\displaystyle \text{angle of inclination,}}\\ {\displaystyle \frac{r}{\ell }}& =& {\displaystyle \text{sin}\hspace{0.17em}\theta ,}\\ {\displaystyle \frac{h}{\ell }}& =& {\displaystyle \text{cos}\hspace{0.17em}\theta \text{.}}\end{array}\end{array}$$ Volume of a cone is $$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi {\left(\ell \text{sin}\hspace{0.17em}\theta \right)}^{2}h=\frac{1}{3}\pi {\left(\ell \text{sin}\hspace{0.17em}\theta \right)}^2\ell \text{cos}\hspace{0.17em}\theta \text{.}$$ $\ell$ is fixed (given slant height). We want to maximize $V$ as $\theta$ changes $$\begin{array}{ccc}{\displaystyle \frac{dV}{d\theta }}& =& {\displaystyle \frac{d\frac{1}{3}\pi {\ell }^3{\text{sin}}^2\theta \text{cos}\hspace{0.17em}\theta }{d\theta }}\\ & =& {\displaystyle \frac{1}{3}\pi {\ell }^3(2\text{sin}\hspace{0.17em}\theta \hspace{0.17em}{\text{cos}}^2\theta -{\text{sin}}^2\theta \hspace{0.17em}\text{sin}\hspace{0.17em}\theta )}\\ & =& {\displaystyle \frac{1}{3}\pi {\ell }^3\text{sin}\hspace{0.17em}\theta (2{\text{cos}}^2\theta -{\text{sin}}^2\theta )\text{.}}\end{array}$$ A critical point is when $\frac{dV}{d\theta }$ is zero or when $2{\text{cos}}^2\theta -{\text{sin}}^2\theta =0$ or $\text{sin}\hspace{0.17em}\theta =0\text{.}$ So $2={\text{tan}}^2\theta$ or $\theta =0\text{.}$
So $\sqrt{2}=\text{tan}\hspace{0.17em}\theta$ or $\theta =0\text{.}$
So $\theta ={\text{tan}}^{-1}\sqrt{2}$ or $\theta =0\text{.}$ When $\theta =0$ the cone looks like $\begin{array}{c} \end{array}$ which clearly does not have maximum volume. So $\theta ={\text{tan}}^{-1}\sqrt{2}$ maximizes volume.

Notes and References

These are a typed copy of Lecture 20 from a series of handwritten lecture notes for the class MATH 221 given on October 25, 2000.

page history