Originally Posted by

**ticbol** Let x = length of cut for the circle

So (10-x) = for the square

Circle:

Circumference = 2pi*r = x

So, r = x/(2pi)

Area = pi(r^2) = pi(x / 2pi)^2 = (x^2)/(4pi) ----**

Square:

Perimeter = 4s = 10-x

So, s = (10-x)/4

Area = s^2 = (100 -20x +x^2)/16 -----***

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a) Combined areas as small as possible.

A = (x^2)/(4pi) +(100 -20x +x^2)/16

dA/dx = 2x/(4pi) +(-20 +2x)/16

Set that to zero,

0 = x/(2pi) +(x-10)/8

Clear the fractions, multiply both sides by 8pi,

0 = 4x +pi(x-10)

0 = 4x +pi(x) -10pi

x = 10pi /(pi+4) = 4.4 ft

So, 10 -x = 5.6 ft

Are those for least A?

Let's check the concavity of the graph of A at x=4.4 ft.

A' = x/(2pi) +(x-10)/8

A'' = 1/(pi) +1/8 = positive, concave upwards. -----> minimum.

Therefore, for least combined areas, cut the 10-ft wire into 4.4 ft and 5.6 ft. -------answer.

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b) For combined areas as large as possible.

Umm. We will go through the exact same route as above for the max combined areas. Same dA/dx. So only for minimum.

What does that mean?

No possible way for maximum combined areas?

Don't know. Beats me.

Or, 4.4 ft and 5.6 ft cuts give the optimum combined areas.