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36_Cal_Solution of Calculus_6e

# 36_Cal_Solution of Calculus_6e - Answer V A(V = 4 1/3 V r=...

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T h x 2 x r = V π 1 / 3 and so A = 4 π V π 2 / 3 . Answer: A ( V ) = 4 π V π 2 / 3 , 0 < V < + . It is permissible, and sometimes desirable, to use instead the domain 0 V < + . C01S0M.017: The following figure shows an equilateral triangle with sides of length 2 x and an altitude of length h . Because T is a right triangle, we see that x 2 + h 2 = (2 x ) 2 , so that h = x 3 . The area of this triangle is A = hx and its perimeter is P = 6 x . So A = x 2 3 and x = P 6 . Therefore A ( P ) = P 2 3 36 , 0 < P < . C01S0M.018: The square has perimeter x and thus edge length y = 1 4 x . The circle has circumference 100 x . Thus if z is the radius of the circle, then 2 πz = 100 x , so that z = (100 x ) / (2 π ). The area of the square is y 2 and the area of the circle is πz 2 , so that the sum of the areas of the square and the circle is
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