It is given that V(x)=350x/(1+x^2)^(3/4). The value of x for which V(x) is maximum has to be determined.
Find the derivative of V(x). And solve for V'(x) = 0.
V'(x) = [350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4]/(1+x^2)^(3/4)
V'(x) = 0
=> [350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4]/(1+x^2)^(3/4) = 0
=> 350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4 = 0
=>...
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It is given that V(x)=350x/(1+x^2)^(3/4). The value of x for which V(x) is maximum has to be determined.
Find the derivative of V(x). And solve for V'(x) = 0.
V'(x) = [350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4]/(1+x^2)^(3/4)
V'(x) = 0
=> [350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4]/(1+x^2)^(3/4) = 0
=> 350*(1+x^2)^(3/4) - 350x*2x*(3/4)(1+x^2)^-1/4 = 0
=> (1 + x^2) - 2x^2(3/4) = 0
=> 2(1 + x^2) - 3x^2 = 0
=> 2 + 2x^2 - 3x^2 = 0
=> x^2 = 2
=> x = sqrt 2
There could be other solutions of x but x = sqrt 2 is the only real and positive solution. x = sqrt 2 corresponds to approximately 14142.3 light years.
The value of V(x) is maximum at x = sqrt 2