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Sagot :
ANSWER:
900
STEP-BY-STEP EXPLANATION:
Given the cost and revenue functions:
[tex]\begin{gathered} C\mleft(x\mright)=-21x^2+52000x+20940 \\ R\mleft(x\mright)=-29x^2+196000x \end{gathered}[/tex]The profit is:
[tex]\begin{gathered} P(x)=R(x)-C(x) \\ P(x)=-29x^2+196000x-(-21x^2+52000x+20940) \\ P(x)=-29x^2+196000x+21x^2-52000x-20940 \\ P(x)=-8x^2+144000x-20940 \end{gathered}[/tex]To maximize we derive the profit function and then set it equal to 0 to solve for x
[tex]\begin{gathered} P^{\prime}(x)=\frac{d}{dx}(-8x^2+144000x-20940) \\ P^{\prime}(x)=-16x+144000 \\ P^{\prime}(x)=0 \\ 0=-16x+144000 \\ 16x=14400 \\ x=\frac{14400}{16} \\ x=900 \end{gathered}[/tex]To maximize profit, a total of 900 pPhones must be produced and sold.
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