A wire of length 10 is cut into two pieces which are then bent into the shape of a circle of radius r and a square of side ss. Then the total area enclosed by the circle and square is the following function of ss and rIf we solve for s in terms of r , we can reexpress this area as the following function of r alone: Thus we find that to obtain maximal area we should let r= To obtain minimal area we should let r=

s = (5 - πr)/2

r = 5/(4(1+π))

r is 5/(4π) to obtain minimal area

Step-by-step explanation:

Given

Length of wire = 10cm

The length of the wire is being divided between the perimeter of the square and the circumference of the circle.

Let s = length of a side of the square.

Let r = Radius of the circle

Perimeter of Square = 4s

Circumference b = 2πr

Perimeter of the Square and circle

4s + 2πr = 10 --- make s the subject of formula

4s = 10 - 2πr

s = (10 - 2πr)/4

s = (5 - πr)/2

Area of the square and circle

πr² + s² = A --- substitute (5-πr)/2 for s

πr² + (5-πr)²/2² = A

πr² + (25 - 10πr + (πr)² )/4= A

A = πr² + (πr)²/4 - 10πr/4 + 25/4

Differentiating with respect to r

dA/dr = 2πr + 2π²r/4 - 10π/4

0 = 2πr + 2π²r/4 - 10π/4

Solve for r

2πr + 2π²r = 10π/4

r(2π + 2π²) = 10π/4

r = 10π/4 / (2π + 2π²)

r = 10π/4 / 2π((1 + π))

r = 5/(4(1+π))

Since A(r) is concave up, it has a minimum at

r = 5/(4(1+π))

s = (5 - πr)/2 and s>0

Since s = (15 - 2πr ) / 4 and s > 0,

the maximum r is 5/(4π).

This same r will also maximize the area.

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6