Saturday

December 20, 2014

December 20, 2014

Posted by **Andy G** on Wednesday, January 22, 2014 at 8:00pm.

a. Write the function that represents the are of the two figures combined.

b. Determine the domain of the function.

c. Find the value(s) of x that yield a max and minimum area.

d. explain your reasoning.

- Pre-Calc -
**Reiny**, Wednesday, January 22, 2014 at 8:59pmlet me say right away, that defining x to be the length of the wire used for the square is a poor choice. But ...

let the piece used for the square be x

then the piece used for the circle = 100-x

side of square = x/4

area of square = x^2 /16

perimeter of circle = 100-x

perimiter of circle = 2πr

2πr = 100-x

r = (100-x)/(2π)

area of circle = π(100-x)^2/(4π^2) = (100-x)^2/(4π)

total area = (1/16)x^2 + (1/(4π))(100-x)^2

domain: clearly 0 < x < 100

Since you labeled your problem as "Pre-Calc" I can't take the derivative and set it equal to zero, like I want to.

So, expanding and simplifying the area equation I got

Area = (1/16 + 1/(4π))x^2 - (50/π)x + 2500/π

which is a parabola opening upwards, the vertex would represent a**minimum**area

the x of the vertex is -b/(2a)

= (50/π) / (2(1/16 + 1/(4π))) = appr 56 cm

so a min total will be obtained when**x = 56 cm**

clearly the maximum will be obtained if x = 100, that is all is used for a circle.

Here is how I would do this:

let the radius be x, and the side of the square be y

4y + 2πx = 100

4y = 100 - 2πr

y = 25 - πx/2

area = πx^2 + (25-πx/2)^2

=πx^2 + 625 - 25πx + π^2 x^2/4

= (π + π^2/4)x^2 - 25πx + 625

the x coordinate of this parabola is 25π/(2π + π^2/2)

= 25/(2 + π/2) = appr 7.00

so the length for the circle = 2πx = 44

and the circle uses 100-44 or 56 cm, same as above.

**Answer this Question**

**Related Questions**

math - A florist uses wire frames to support flower arrangements displayed at ...

Calc - A wire 9 meters long is cut into two pieces. One piece is bent into a ...

MINIMIZATION PROBLEM (CALC) - A wire 9 meters long is cut into two pieces. One ...

MINIMIZATION PROBLEM (CALC) - A wire 9 meters long is cut into two pieces. One ...

Calculus - A piece of wire 40 m long is cut into two pieces. One piece is bent ...

Calculus - A wire 4 meters long is cut into two pieces. One piece is bent into a...

Calc - A wire 100 inches long is to be cut into two pieces. One of the pieces ...

calculus - A five feet piece of wire is cut into two pieces. One Piece is bent ...

calc and Vectors - A wire 60 cm long is to be cut into two pieces. One of the ...

Calculus - A piece of wire 12 m long is cut into two pieces. One piece is bent ...