A piece of wire 5 inches long is to be cut into two pieces. One peice is x inches long and is to be bent into the shape of a square. The other piece is to be bent into the shape of a circle. Find an expression for the total adrea made up by the square and the circle as a function of x.

Question

campbell_st · Answer

for the square

Perimeter = x

side length = x/4

then area   A = x^2/16

The circle   if x inches in used there is 5 - x inches remaining

so the circumference = 5 -x

find the radius

$$5 - x = 2\pi r$$

$$r = \frac{5 - x}{\pi}$$

area of the circle 

$$A = \pi \times(\frac{5 -x}{\pi})^2$$

$$A = \frac{1}{\pi} (5 - x)^2$$

or 

$$A = \frac{1}{\pi} 25 - 10x + x^2$$

then the total area is 

$$f(x) = \frac{x^2}{16} + \frac{1}{\pi}( 25 - 10x + x^2)$$


giving the final function as 

$$f(x) = (\frac{1}{16} + \frac{1}{\pi})x^2 -10x + 25$$

campbell_st · Answer

you many want to simplify the coefficient of x^2

anonymous · Answer

what happened to the 2 when you divided 2π to find r?

campbell_st · Answer

oops missed a 2 

the radius should be 

$$r = \frac{5 -x}{2\pi}$$

so area of the circle is 

$$A = \pi (\frac{(5 -x)}{2\pi})^2$$

$$A = \frac{(5 - x)^2}{4\pi}$$

so the total area is 

$$f(x) = \frac{x^2}{16} + \frac{(5 - x)^2}{2\pi}$$

I'll let you simplify it... sorry for the error in the 1st posting

campbell_st · Answer

dang... total area

$$f(x) = \frac{x^2}{16} + \frac{(5 -x)^2}{4\pi}$$

which is finally correct... sorry for the typos

anonymous · Answer

thanks!

anonymous · Answer

close this ..