Answer :
Solution:
Given:
[tex]\begin{gathered} C(x)=x^3-886.5x^2-10983x+1651 \\ P(x)=21x+15 \end{gathered}[/tex]The revenue is given by;
[tex]\begin{gathered} R(x)=P(x).x \\ R(x)=(21x+15)x \\ R(x)=21x^2+15x \end{gathered}[/tex]Hence, the profit is the difference between the revenue and the cost.
Thus,
[tex]\begin{gathered} Profit=R(x)-C(x) \\ Profit=21x^2+15x-(x^3-886.5x^2-10983x+1651) \\ Profit=21x^2+15x-x^3+886.5x^2+10983x-1651 \\ Profit=-x^3+907.5x^2+10998x-1651 \\ Let\text{ profit be represented by G\lparen x\rparen.} \\ Hence,\text{ } \\ G(x)=-x^3+907.5x^2+10998x-1651 \end{gathered}[/tex]To get the maximum profit, we differentiate the profit function and equate it to zero.
Hence,
[tex]\begin{gathered} G(x)=-x^3+907.5x^2+10998x-1651 \\ G^{\prime}(x)=-3x^2+1815x+10998 \\ when\text{ }G^{\prime}(x)=0, \\ 0=-3x^2+1815x+10998 \\ Rewrite\text{ the equation;} \\ 3x^2-1815x-10998=0 \\ 3(x^2-605x-3666)=0 \\ x^2-605x-3666=0 \\ \\ \\ \\ \\ Factorizing\text{ the quadratic equation,} \\ x=611\text{ OR }x=-6 \end{gathered}[/tex]To get the maximum production level, we differentiate further.
[tex]\begin{gathered} G^{\prime}(x)=-3x^{2}+1,815x+10,998 \\ G^{\prime}^{\prime}(x)=-6x+1815 \\ \\ when\text{ }G^{\prime}^{\prime}(x)<0,\text{ it is maximum, otherwise it is minimum.} \\ \\ Hence,\text{ at x = 611,} \\ G^{\prime}^{\prime}(x)=-6(611)+1815=-1851 \\ \\ Hence,\text{ maximum occurs at x = 611} \end{gathered}[/tex]Therefore, the production level that will maximize profit is x = 611.
