MHB Chelsea's question at Yahoo Answers regarding functions and optimization

[MarkFL]

Here is the question:

I need help on a HW problem - I got the first two parts right, but am having trouble finishing the last two?

I need help on a HW problem - I got the first two parts right, but am having trouble finishing the last two parts.

The figure shows a sector with radius r and angle θ in radians. The total perimeter of the sector is 120 cm.

(a) Express θ as a function of r.

I got
Theta = 2r-120/-r

(b) Express the area of the sector as a function of r.
I got A=1/2 * r^2 * (2r-120/-r)

Now these are the two that I am having trouble figuring out.

(c) For what radius r is the area a maximum?
in cm
and

(d) What is the maximum area?
in cm^2I have to answer them as exact answers - no decimals allowed. Is there anyone who would be willing to walk me through this?

Thanks.

Here is a link to the question:

I need help on a HW problem - I got the first two parts right, but am having trouble finishing the last two? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[MarkFL]

Hello Chelsea,

You have made a minor error in part (a), so let's work this problem from the beginning. All linear measures will be in cm.

(a) Express $\theta$ as a function of $r$.

From the information regarding the perimeter of the sector, we may state the following:

$$2r+r\theta=120$$

Now, we wish to solve for $\theta$. We may begin by subtracting through by $2r$ to get:

$$r\theta=120-2r$$

Now, we may divide through by $r$ to get:

$$\theta=\frac{120}{r}-2$$

Thus, we have expressed $\theta$ as a function of $r$.

(b) Express the area of the sector as a function of $r$.

The area $A$ of a circular sector is given by:

$$A=\frac{1}{2}r^2\theta$$

Now we merely need to substitute for $\theta$, and simplify:

$$A=\frac{1}{2}r^2\left(\frac{120}{r}-2 \right)=60r-r^2$$

Thus, we have expressed $A$ as a function of $r$.

(c) For what radius $r$ is the area a maximum?

We could observe that the area function is a quadratic in $r$, with the roots $r=0,\,60$ and thus the axis of symmetry must be the line $r=30$ which is where the vertex is. Since the area function is a parabola opening down, we know the vertex is at the global maximum, and so $A$ is maximized for $r=30$.

You may however be expected to use differential calculus to maximize the function, and so we find by differentiating with respect to $r$ and equating to zero:

$$A'(r)=60-2r=0\,\therefore\,30-r=0\,\therefore\,r=30$$

Now since we have $$A''(r)=-2<0$$, we know by the second derivative test that the extremum we have found is a maximum.

(d) What is the maximum area? in cm^2

To find the maximum area, we let $r=30$ in the area function:

$$A_{\text{max}}=A(30)=60(30)-30^2=30^2(2-1)=900\text{ cm}^2$$

To Chelsea and any other guests viewing this topic, I would invite and encourage you to post other optimization problems in our http://www.mathhelpboards.com/f10/ forum.

Best Regards,

Mark.