Steve likes to entertain friends at parties with "wire tricks." suppose he takes a piece of wire 72 inches long and cuts it into two pieces. steve takes the first piece of wire and bends it into the shape of a perfect circle. he then proceeds to bend the second piece of wire into the shape of a perfect square. what should the lengths of the wires be so that the total area of the circle and square combined is as small as possible? (round your answers to two decimal places.)

The solution for this problem is:

Cut wire so first piece has length x. 
Second piece has length (72 - x). 

Use piece of length x to make circle. 
c = circumference, r = radius 
c = 2πr = x 
r = x/(2π) 

A(circle) = πr² = π * (x/(2π))² = x²π/4π² = x²/4π 

Use piece of length (72-x) to make square. 
s = side length = (72-x)/4 

Area(square) = s² = ((72-x)/4)² = (72-x)²/16 = (5184 - 144x + x²)/16 
Area(square) = 324 – 9x + x²/16 

A = A(circle) + Area(square) 
A = x²/4π + 324 - 9x + x²/16 
A = x²/4π + x²/16 – 9x + 324 
A = 4x²/16π + πx²/16π - 9x + 324
A = (4+π)/16π x² - 9x + 324 

This is the function of a parabola that opens up. 

To look where A is minimum, you can rewrite equation in vertex form or find where derivative = 0. 

A' = 2(4+π)/16π x - 9 = (4+π)/8π x - 9 
A' = 0 
(4+π)/8π x - 9 = 0 
(4+π)/8π x = 9 
x = 9*8π / (4+π) 
x ≈ 31.7 inches 

3x^3-x^2+6x-9

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length