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July 23rd, 2015, 11:59 PM   #1
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Question 1/4 circle and rectangle - Differentiation - Calculus

The cross-section of an object has the shape of a quarter-circle of radius r adjoining a rectangle of width x and height r. As shown in the diagram (attached).

a) The perimeter and area of the cross-section are P and A respectively. Express each of P and A in terms of r and x, and hence show that
A=1/2Pr - r^2.

This I worked out, no trouble:
P=2πr/4 + 2r + 2x = 2πr/2 + 2r + 2x

A= πr^2/4 + xr
x= A/r - πr/4 Put this into P

P= 2r + 2A/r
Pr= 2r^2 + 2A
Pr/2= r^2 + A
A=pr/2 - r^2

b) Talking the perimeter P of the cross-section as fixed, find x in terms of r for the case when the area A of the cross-section is a maximum, and show that, for this value of x, A is a maximum and not a minimum.

I have no idea where to start..
Thanks!
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July 24th, 2015, 01:01 AM   #2
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Find $\dfrac{dA}{dr}$, set it equal to 0 and solve for $r$. This will give you a critical point. To show that it is a maximum, either investigate the sign of $\dfrac{d^2A}{dr^2}$ or note that $A$ is a concave down parabola.

The rest is fairly simple.
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July 24th, 2015, 02:16 AM   #3
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So, dA/dr = pr/2 - r^2
What about P?
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July 24th, 2015, 03:08 AM   #4
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You have $A = \dfrac{Pr}{2} - r^2$. If $P$ is fixed, we can treat it like a constant. So $\dfrac{dA}{dr} = \dfrac{P}{2} - 2r$.

Now, set $\dfrac{dA}{dr} = 0$ and solve for $r$.

$\dfrac{P}{2} - 2r = 0$

$2r = \dfrac{P}{2}$

$r = \dfrac{P}{4}$

So we know there is a critical point of $A$ at $r = \dfrac{P}{4}$. Since the graph of $A$ is a concave down parabola, the critical point must be a maximum.

To find the maximum value of $A$, we plug in $r = \dfrac{P}{4}$ into our equation for $A$.

$A_{\text{max}} = \dfrac{P^2}{8} - \dfrac{P^2}{16} = \dfrac{P^2}{16}$

But you have shown another formula for $A$:

$A = \dfrac{\pi r^2}{4} + xr$

At the point where $A$ is maximum (that is, $A = A_{\text{max}}$) we have.

$\dfrac{\pi r^2}{4} + xr = \dfrac{P^2}{16}$

Now find $x$ in terms of $r$.
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July 24th, 2015, 05:13 AM   #5
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Okay great! I understand everything. The part to find x in terms of r though, the answer in the book is x=1/4(r(4-π))
I almost get there, but something i'snt right. The P in that last equation, am i meant to use P=4r?

I get to x=(1-πr)/4 using that. which simplifies to 1/4(4-πr) obviously. But thats not right!

Last edited by Jmun; July 24th, 2015 at 05:18 AM.
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July 24th, 2015, 08:29 PM   #6
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Using $P = 4r$ is correct. Check your algebra again. I seem to get the right answer.
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July 25th, 2015, 03:26 AM   #7
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Talking

okay great! thanks so much!
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