Probem41+Julie,+Marnie

Problem 41 - Marnie and Julie

Our objection funtion is P = (E²R) / (R + r)² and we are maximizing it

E and r are constants, R is the only variable

Lim P = 0 R→ ∞

Lim P = 0 R→ -∞  The derivative of the function is: P'=[ E²(R+r)² - 2E²R(R+r) ]/ (R+r)⁴

0= (R+r)^2(E^2)-2(E^2)R(R+r) 0= (E^2)(R+r)[R+r - 2R] 0= (E^2)(R+r)(r-R)

(cancel out E^2 b/c it is a constant)

so, R= -r and R=r

(R cannot equal -r so cancel that out)

R=r is the critical value

P(r) = (E^2)r/(2r)^2 = (E^2)/4r

lim as P approaches 0 is P = 0

lim as P approaches infinity is P = 0

The answer is (E^2)/4r

Set P' equal to zero to find the critical values:

[E²R+E²r - 2E²R] / (R+r)³ = 0

E²R+E²r - 2E²R = 0

E²R - 2E²R = -E²r

R(E² - 2E²) = -E²r

R = -E²r / (E² - 2E²)

R = - [ -E²r / (E² - 2E²) ]

R = E²r / ( 2E² - E² )

R = r / (2E² - 1)

P'(0) = 0, r / (2E² - 1) P [r / (2E² - 1)] = E² / 4r

Answer = E² / 4r