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July 23rd, 2015, 11:59 PM  #1 
Newbie Joined: Jun 2015 From: South Africa Posts: 27 Thanks: 0  1/4 circle and rectangle  Differentiation  Calculus
The crosssection of an object has the shape of a quartercircle 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 crosssection 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 crosssection as fixed, find x in terms of r for the case when the area A of the crosssection 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! 
July 24th, 2015, 01:01 AM  #2 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
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. 
July 24th, 2015, 02:16 AM  #3 
Newbie Joined: Jun 2015 From: South Africa Posts: 27 Thanks: 0 
So, dA/dr = pr/2  r^2 What about P? 
July 24th, 2015, 03:08 AM  #4 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
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 $r$. 
July 24th, 2015, 05:13 AM  #5 
Newbie Joined: Jun 2015 From: South Africa Posts: 27 Thanks: 0 
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(4π)) 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=(1πr)/4 using that. which simplifies to 1/4(4πr) obviously. But thats not right! Last edited by Jmun; July 24th, 2015 at 05:18 AM. 
July 24th, 2015, 08:29 PM  #6 
Math Team Joined: Nov 2014 From: Australia Posts: 689 Thanks: 244 
Using $P = 4r$ is correct. Check your algebra again. I seem to get the right answer.

July 25th, 2015, 03:26 AM  #7 
Newbie Joined: Jun 2015 From: South Africa Posts: 27 Thanks: 0  okay great! thanks so much!


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1 or 4, calculus, circle, differentiation, rectangle 
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