Question

3. Find the angle of the largest right circular cone which can be inscribed in a sphere of

radius 9 inches.

4. A statue 10 feet high is standing on a base 13 feet high. If an observer’s eye is 5 feet

above the ground, how far should he stand from the base in order that the angle

between his lines of sight to the top and bottom of the statue is a maximum. (How far

should he stand to get the best view of the statue.

5. A steel girder 27 feet long is to be moved on rollers along a passageway 8 feet in

width and into a corridor at right angles to the passageway. If the horizontal width of

the girder is neglected, how wide must the corridor be in order that the girder can go

around the corner?

Answer 1

3.

Let $R=$ the radius of the sphere.

Let $AD=r, BD=h, \angle ABC=\theta.$

$\triangle ABC$

$AC=2r, AO=OC=R, \angle AOC=2\angle ABC=2\theta$

The Law of Cosines

$$(2r)^2=R^2+R^2-2R^2\cos(2\theta)$$

$$4r^2=2R^2(2\sin^2\theta)$$

$$r=R\sin \theta$$

$$h=r\tan(\theta/2)=R\sin \theta\tan(\theta/2)$$

$$V_{cone}=\dfrac{1}{3}\pi r^2 h$$

$$V_{cone}=V_{cone}(\theta)=\dfrac{1}{3}\pi R^3\sin^3 \theta\tan(\theta/2)$$

$$(V_{cone})'_{\theta}=\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos(\theta)\tan(\theta/2)$$

$$+\dfrac{1}{2}\sin^3\theta(\dfrac{1}{\cos ^2(\theta/2)})\bigg)$$

Find the critical numbr(s)

$$(V_{cone})'_{\theta}=0$$

$$\dfrac{1}{3}\pi R^3\bigg(3\sin^2 \theta\cos\theta\tan(\theta/2)$$

$$+\dfrac{\sin^3\theta}{2\cos ^2(\theta/2)}\bigg)=0$$

$$\dfrac{\sin^3 \theta(3\cos\theta+1)}{2\cos ^2(\theta/2)}=0$$

$$\cos \theta=-\dfrac{1}{3}$$

If $0< \theta<\pi-\cos^{-1}(1/3), (V_{cone})'_{\theta}>0, V_{cone}$ increases.

If $\pi-\cos^{-1}(1/3)<\theta<\pi, (V_{cone})'_{\theta}<0, V_{cone}$ decreases.

The volune of inscribed cone has the absolute maximum at

$$\theta=\pi-\cos^{-1}(1/3)$$

4.

$$\tan \alpha=\dfrac{13-5}{x}=\dfrac{8}{x}$$

$$\tan (\alpha+\theta)=\dfrac{13-5+10}{x}=\dfrac{18}{x}$$

$$\tan (\alpha+\theta)=\dfrac{\tan \alpha+\tan \theta}{1-\tan \alpha\tan \theta}$$

$$\dfrac{\dfrac{8}{x}+\tan \theta}{1-\dfrac{8}{x}(\tan \theta)}=\dfrac{18}{x}$$

$$8+x\tan \theta=18-\dfrac{144}{x}(\tan \theta)$$

$$\tan \theta=\dfrac{10x}{x^2+144}$$

$$(\tan\theta)'_x=\dfrac{10(x^2+144-2x^2)}{(x^2+144)^2}$$

$$=\dfrac{10(144-x^2)}{(x^2+144)^2}$$

Find the critical number(s)

$$(\tan\theta)'_x=0=>\dfrac{10(144-x^2)}{(x^2+144)^2}=0$$

$$x_1=-12, x_2=12$$

We consider $x\geq 0$

If $0\leq x<12, (\tan\theta)'_x=0>0, \tan \theta$ increases.

If $x>12, (\tan\theta)'_x=0<0, \tan \theta$ decreases.

The angle $\theta$ has the local maximum at $x=12$ ft.

Since the function $\tan \theta$ has the only extremum for $x\geq 0,$ then the angle $\theta$ has the absolute maximum for $x\geq 0$ at $x=12$ ft.

5.

$$a=\dfrac{8}{\cos \theta}, b=\dfrac{w}{\sin \theta}$$

$$a+b=27$$

$$\dfrac{8}{\cos \theta}+\dfrac{w}{\sin \theta}=24$$

$$w=27\sin \theta-8\tan \theta$$

Find the first derivative

$$w'_{\theta}=27\cos \theta-\dfrac{8}{\cos^2\theta}$$

Find the critical number(s)

$$w'_{\theta}=0=>27\cos \theta-\dfrac{8}{\cos^2\theta}=0$$

$$27\cos^3\theta=8$$

$$\cos \theta=2/3$$

$$\theta=\cos^{-1}(2/3)$$

If $0<\theta<\cos^{-1}(2/3), w'_{\theta}>0, w$ increases.

If $\cos^{-1}(2/3)<\theta<\pi/2, w'_{\theta}<0, w$ decreases.

$$\sin^2\theta=1-(2/3)^2=5/9$$

$$\sin \theta=\sqrt{5}/3$$

$$w=27(\sqrt{5}/3)-8(\sqrt{5}/2)$$

$$w=5\sqrt{5}\ ft$$