Probem34+Jimmy,+Carolyn

x+y=10 y=10-x

square: x/4=side (s) circle: 2πr =y

r =(10-x)/2 π
Optimize:(do you mean maximize or minimize?)

A=s^2 + πr^2

A=(x/4)^2 + π (y/2 π)^2
. First find the critical values (need to eventually find the max and min): A= (x/4)^2 + π( (10-x)/2 π)^2 A'= 2(x/4) (1/4) + π (2((10-x)/2 π))(-1/2 π) A'= .5(x/4) + π (- π(10-x)/(2 π)) A'= x/8 + π(-(10-x)/2) (ms sweeney: I still think you lost a pi in the denominator!) A'= x/8 + (-10 π+ πx )/2 A'=x/8 -5 π+( π/2)x 0= x/8 -5 π+( π/2)x 5 π=(1/8)x + ( π/2)x 5 π=((4 π+1)/8 )x x=9.263 A(9.263)= 5.406

Limits: lim A(x) as x-->0 = π (10/2 π)^2 = 7.958 lim A(x) as x-->10= 6.25

Therefore the maximum occurs as x approaches 0 (I do not think x can equal 0.. can it?). The minimum occurs at x=9.263 And there's your problem. lim A(x) as x-->infinity= infinity CANNOT exist because of the domain of the function which is [0,10]. At no point along the domain, does A'(x) = 0, so you must look at the end points. So the maximum area occurs at x=0. So the maximum area is 25 / π with the circumference of the square equaling 0 and the circumference of the circle equaling 10.