Maxima and Minima problems - help please :) (1 Viewer)

swifty13 · Apr 19, 2012

Hi I am having trouble with 2 max and min problems if anyone would help me with them it'd be much appreciated

1. A piece of wire 10 metres long is cut into two portions. One piece is bet to form a cricle and the other piece to form a square. FInd the circumference of the circle if the sum of areas of the circle and square is to be minimum. Give your answer in terms of pi.

2. A rectangular crate is required to have a volume of 24 cubic metres and its length is to be double its width. Also it is to be strengthened by a strip of angle iron running along all its edges. Find the dimensions of the crate if the total length of angle iron is to be minimum.

So far all I have for the second question is P = 16x, I labelled the length as 2x and the width as x and added it together. I'm a bit confused cause I know I should have to variables in my equation like an x and a y but I only have one as i'm not sure what I'm looking for.

nightweaver066 · Apr 19, 2012

$1. Let the circumference of the circle be x and the perimeter, P, of the square be 10 - x$

$\therefore 2\pi r = x ~ , ~ P = 10 - x$

$r = \frac{x}{2\pi} ~ , ~ s = \frac{10 - x}{4} ~ $where s is the side length of the square$$

$\therefore A_c = \pi \times (\frac{x}{2\pi})^2 ~ , ~ A_s = \frac{(10 - x)^2}{16}$

$A_c + A_s = \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16}$

$\frac{d}{dx}(A_c + A_s) = \frac{x}{2\pi} - \frac{(10 - x)}{8}$

$$For max/min sum of areas, let $ \frac{d}{dx}(A_c + A_s) = 0$

$4x - \pi (10 - x) = 0$

$4x - 10\pi + \pi x = 0$

$x = \frac{10\pi}{\pi + 4}$

$Test for maximum/minimum using table method or second derivative and you should get a minimum..$

jmadden · Apr 20, 2012

nightweaver066 said:
$1. Let the circumference of the circle be x and the perimeter, P, of the square be 10 - x$

$\therefore 2\pi r = x ~ , ~ P = 10 - x$

$r = \frac{x}{2\pi} ~ , ~ s = \frac{10 - x}{4} ~ $where s is the side length of the square$$

$\therefore A_c = \pi \times (\frac{x}{2\pi})^2 ~ , ~ A_s = \frac{(10 - x)^2}{16}$

$A_c + A_s = \frac{x^2}{4\pi} + \frac{(10 - x)^2}{16}$

$\frac{d}{dx}(A_c + A_s) = \frac{x}{2\pi} - \frac{(10 - x)}{8}$

$$For max/min sum of areas, let $ \frac{d}{dx}(A_c + A_s) = 0$

$4x - \pi (10 - x) = 0$

$4x - 10\pi + \pi x = 0$

$x = \frac{10\pi}{\pi + 4}$

$Test for maximum/minimum using table method or second derivative and you should get a minimum..$

How do you insert maths working into comments like this?

RealiseNothing · Apr 20, 2012

jmadden said:
How do you insert maths working into comments like this?

[tex.] insert maths here [/tex.]

Without the full stops.

Use this site if you need to learn the functions:

http://www.codecogs.com/latex/eqneditor.php

qwerty44 · Apr 20, 2012

qwerty44 · Apr 20, 2012

lol ^ not very aligned

Maxima and Minima problems - help please :) (1 Viewer)

swifty13

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RealiseNothing

what is that?It is Cowpea

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