A piece of wire of length 60 is cut, and the resulting two pieces are formed to make a circle and a square. Where should the wire be cut to (a) minimize and (b) maximize the combined area of the circle and square?
So Area of circle and square= pir^2 +x^2 perimeter of circle and square = 2pir+4x
Solve for r: 2pir+4x=60 r=30-2x/pi
plug r into area equation
pi(30-2x/pi)^2+x^2=area (30-2x)^2/pi + x^2=
find derivative to get CP for min/max
A(X)=(30-2x)^2/pi + x^2 A'(x)= (-120+2x/pi)+2x
therefore, x=60 and x=0.
Now at this point I know that if I plug in 60 and 0 in the original area function, it will give me a number then I can compare those numbers and see where is it largest at what x value. But how would I find the minimum value exactly?
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$\begingroup$You made a mistake in final calculations. You have the area as a function of $x$: $$ A(x)=(30-2x)^2/\pi+x^2 $$ So the derivative $$ A'(x)=-120/\pi+8x/\pi+2x $$ $$ A'=0 \to x=60/(\pi+4) $$ This is point of minimum since $A(x)$ is a parabola looking up. Maximum is obtained on 1 of edges $x=0,60$ which is easy to check. Obviously $x=60$ is a maximum. Plug corresponding $x_{min}$ and $x_{max}$ to find maximum/minimum values of the area.
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