My Math Forum 1/4 circle and rectangle - Differentiation - Calculus

 Pre-Calculus Pre-Calculus Math Forum

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 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!
Attached Images
 p.png (2.7 KB, 15 views)

 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(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.
 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!

 Tags 1 or 4, calculus, circle, differentiation, rectangle

,

,

,

,

,

,

# A plate of cookies on the right has the shape of a circle.The radius(r)of this plate is 8 cm as shown onn the picture.calculate the perimeter of the plate

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post link107 Calculus 2 May 4th, 2013 06:09 PM rohit99 Calculus 1 January 26th, 2012 07:08 PM andirrashi Algebra 3 October 24th, 2011 07:38 AM mhurkman Algebra 4 February 7th, 2011 10:40 AM sel Calculus 1 November 25th, 2008 01:57 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top