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The Pear company sells pPhones. The cost to manufacture 2 pPhones isC(x) 19x² + 61000x + 18324 dollars (this includes overhead costs and production costs for eachpPhone). If the company sells 2 pPhones for the maximum price they can fetch, the revenue function willbe R(x) = - 23x² + 1490002 dollars.How many pPhones should the Pear company produce and sell to maximimze profit? (Remember thatprofit=revenue-cost.)x=

Respuesta :

Since the cost function is

[tex]C(x)=-19x^2+61000x+18324[/tex]

Since the revenue function is

[tex]R(x)=-23x^2+149000x[/tex]

Since the profit function is

[tex]P(x)=R(x)-C(x)[/tex]

Then we will subtract C from R

[tex]\begin{gathered} P(x)=-23x^2+149000x-(-19x^2+61000x+18324) \\ P(x)=-23x^2+149000x+19x^2-61000x-18324 \end{gathered}[/tex]

Add the like terms

[tex]\begin{gathered} P(x)=(-23x^2+19x^2)+(149000x-61000x)-18324 \\ P(X)=-4x^2+88000x-18324 \end{gathered}[/tex]

Now, we will differentiate P(x) and equate the answer by 0 to find x maximum

[tex]\begin{gathered} P^{\prime}(x)=-4(2)x^{2-1}+88000(1)x^{1-1} \\ P^{\prime}(x)=-8x+88000 \end{gathered}[/tex]

Equate P' by 0 to find x

[tex]\begin{gathered} P^{\prime}(x)=0 \\ -8x+88000=0 \end{gathered}[/tex]

Subtract 88000 to both sides

[tex]\begin{gathered} -8x+88000-8000=0-88000 \\ -8x=-88000 \end{gathered}[/tex]

Divide both sides by -8

[tex]\begin{gathered} \frac{-8x}{-8}=\frac{-88000}{-8} \\ x=11000 \end{gathered}[/tex]

The company should produce 11000 Phones to maximize the profit