# perimeter and area

• Aug 18th 2010, 01:10 AM
furor celtica
perimeter and area
the cross-section of an object has the shape of a quarter-circle of radius r adjoining a rectangle ofwidth x and height r as shown in the diagram. 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=0.5Pr - r^2 B. taking the perimeter 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. ok the part A. was on problem. but part B. i kinda dont even understand what they are saying, perimeter 'as fixed'? i dont get. i don't understand how to logically combine the given equations, can someone help me wokr it through?
• Aug 18th 2010, 05:56 AM
sa-ri-ga-ma
Perimeter P = πr/2 + 2r + 2x.

Area of cross section A = π*r^2/4 + xr

Substitute the value of x from the above equation. Find dA/dr and equate it to zero. Hence solve for r.
• Aug 18th 2010, 03:50 PM
Quote:

Originally Posted by furor celtica
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.

(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=0.5Pr - r^2

(B). taking the perimeter 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.

ok the part A. was no problem, but part B ? i kinda dont even understand what they are saying, perimeter 'as fixed'? i dont get. i don't understand how to logically combine the given equations, can someone help me work it through?

If the external perimeter is fixed, then if you increase r, x must decrease.
This is because P requires both x and r and if you change one,
you must change the other also, in order to keep P constant.

$P=\frac{2{\pi}r}{4}+2x+2r$ contains x and r.

$A=\frac{{\pi}r^2}{4}+xr=0.5Pr-r^2=-r^2+kr$

since P is a constant.
The graph of A versus r is a quadratic.
It's highest power of r co-efficient is negative.
Therefore the graph is an inverted U-shape.
This means it's derivative is zero only at the maximum,
which is mid-way between the two roots (A=0).

You may be able to finish up.
If not, get back on this.