Derivative, some difficulties.

**Attack** · December 17th 2008, 11:02 AM

Can someone please help me to solve this two problems? I've been sitting for hours trying to understand them.

Here it goes.

A 30 cm string was cut in two pieces. One string was formed as a circle and the second string was shaped as a quadrat.
Show that the sum of the circle and the quadrat's area always exceed 30 cm² regardless where the cut is made.

Found a third problem:

A factory are going to to make cylindershaped cans. The sum of the diameter and height are 80 cm. which measures gives the greatest volume?

Please help me with this!
/Attack.

**JD-Styles** · December 17th 2008, 11:24 AM

.

**running-gag** · December 17th 2008, 11:25 AM

Hi

Let x be the length of the string shaped as a circle
x is the perimeter of the circle
The radius is $r = \frac{x}{2\pi}$
The area is $A_c = \pi r^2 = \pi \frac{x^2}{4\pi^2} = \frac{x^2}{4\pi}$

The length of the other part of the string is 30-x
30-x is the perimeter of the square
The length of one side is $a = \frac{30-x}{4}$

The area is $A_s = a^2 = \frac{(30-x)^2}{16}$

The total area is $A(x) = A_c + A_s = \frac{x^2}{4\pi} + \frac{(30-x)^2}{16}$

Just study the function A(x)

EDIT : beaten again ! Arghh ....

**Attack** · December 17th 2008, 12:05 PM

Well, thanks anyway, it was worth a shot.

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