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October 24, 2014

October 24, 2014

Posted by **Kay** on Wednesday, November 23, 2011 at 5:35pm.

(a) How much of the wire should go to the square to maximize the total area enclosed by both figures?

(b) how much of the wire should go to the square to minimize the total area enclosed by both figures?

- HELLP!! calculus -
**Steve**, Wednesday, November 23, 2011 at 7:27pmlet the length of side of the square be x

Now, the area of a circle is pi r^2, but we have a circumference c = 2pi*r

r = c/2pi and c = (25-4x)

a = x^2 + pi * [(25-4x)/2pi]^2

= x^2 + (25-4x)^2/4pi

This function has a minimum at x = 25/(4 + pi) = 3.5

area is 21.88

but has no global maximum.

At x=0 the area is just the circle = pi * (25/2pi)^2 = 625/4pi = 49.73

At x= 25/4 the area is just the square = 625/16 = 39.06

- HELLP!! calculus -
**Reiny**, Wednesday, November 23, 2011 at 8:24pmAlternate solution

b)

Let the radius of the circle be r

the we need 2πr for the circle

leaving 25-2πr for the square, then each side of the square is

(25-2πr)/4

Area = πr^2 + (25-2πr)^2 /16

= πr^2 + 625/16 - (100/16)πr + (4π^2)/16 r^2

this will graph as a parabola opening upwards, so it will have a minimum area.

d(area)/dr = 2πr - 100π/16 + π^2 r/2

= 0 for a max of area

2πr + π^2 r/2 = 100π/16

times 16 , then solving for r

32πr + 8π^2 r = 100π

r(32π + 8π^2) = 100π

r = 100/(32+8π) = 25/(8+2π) appr. 1.7503

then the circle needs 2πr = appr. 10.998

leaving 25 – 10.998 or 14.003 m for the square.

Plugging r = 1.7503 into my area equation gives

Minimum Area = 21.879

test by taking a value of r slightly above and slightly below r = 1.7503

if r = 1.7 , area = 21.893

if r = 1.7503 , area = 21.879

if r = 1.8 , area = 21.893

a) The maximum area is obtained when all the wire is used for the circle and NONE is used for the square

i.e. , max area = (12.5^2)π = 490.87

- HELLP!! calculus -
**Kay**, Wednesday, November 23, 2011 at 9:43pmhow is the derivative equal to that? shouldn't the last part (ð^2 r/2)be pi*2r/2?

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