Find the maximum profit and the number of units that must be produced and sold in order to yield the maximum profit. Assume that revenue, Upper R left parenthesis x right parenthesisR(x), and cost, Upper C left parenthesis x right parenthesisC(x), are in thousands of dollars, and x is in thousands of unitsR(x)=6x-2x^2 , C(x)=x^3-3x^2+4x+1The production level for the maximum profit is about units.

The production level for the maximum profit is about 1215 units.

Explanation:

The income work is given by R(x)= 6x-2x2 and the cost work is given by C(x)= x3-3x2+4x+1 where x is the quantity of units created and sold.  

At that point the benefit work is P(x) = R(x)- C(x) = (6x-2x2) – (x3-3x2+4x+1) = - x3+x2+2x - 1.  

The benefit will be most extreme when dP/dx is 0 and d2P/dx2 is negative. Here, dP/dx = - 3x2+2x+2 and d2P/dx2 = - 6x+2 .  

On utilizing the quadratic recipe, on the off chance that dP/dx = 0, at that point x = [ - 2 ± √{ 22-4*(- 3)*2]/2*(- 3) = [-2 ± √(4+24)]/(- 6) = (2± √ 28)/6 = (1 ± √7)/3 . Since x can't be negative, consequently x = (1 + √7)/3 = 1.215250437 , state 1.215 ( on adjusting to the closest thousandth).  

Likewise, when x = 1.215, at that point d2P/dx2 is negative.  

Subsequently, the benefit will be greatest when 1215 units are created and sold.  

The creation level for the greatest benefit is around 1215 units.

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