then differentiate.. . . set the derivative to zero to find the value of x that maximizes R.-what you are asked is the get the maximum value of the revenue which is calculated by1/2 x + (40-5x)/(10-x) (this is 1/2 x + y)you can take its first derivative and then let the first derivative to be 0 and solve for xthe answer is x=5.......
Revenue(R) = quantity of regular steel * ($S/2) + quantity of super steel *($S)
= x * S/2 + 40-(5x/10-x) * S
you can assume the price S = $2 and S/2 = $1
For y is it y= 40-(5x/10-x) or (40-5x)/(10-x)?
But do you think you can solve now?
what you are asked is the get the maximum value of the revenue which is calculated by
1/2 x + (40-5x)/(10-x) (this is 1/2 x + y)
you can take its first derivative and then let the first derivative to be 0 and solve for x
the answer is x=5.53
