A product can be produced at a total cost (in dollars) of C (x )equals 0.05 x squared plus 10 x plus 1100, where x is the number of units produced. If the total revenue (in dollars) is given by R (x )equals 46 x minus 0.01 x squared, determine the number of units required to maximize the profit. Round your answer to 2 decimal places, if necessary. [A] units

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okpalawalter8 okpalawalter8 · Answer 1 · 2021-07-19T06:13:36+02:00

Answer:

[tex]Max\ Profit=300units[/tex]

Step-by-step explanation:

From the question we are told that:

[tex]C(x)=0.05x^2+10x^2+1100[/tex]

[tex]R(x)=4.6x-0.01x^2[/tex]

Generally the equation for Profit is is mathematically given by

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

[tex]P(x)=(4.6x-0.01x^2)-(0.05x^2+10x^2+1100[/tex]

[tex]P(x)=36x-0.06x^2-1100[/tex]

Differentiate P(x) we have

[tex]P'(x)=36-2(0.06)x[/tex]

Equating to 0

[tex]P'(x)=0[/tex]

[tex]36-2(0.06)x=0[/tex]

[tex]x=300units[/tex]

Second order differentiation, we have

[tex]P''(x)=-0.12[/tex]

Therefore

This implies that

[tex]P''(300)<0[/tex]

Hence, Maximum Profit is actualized at

[tex]Max\ Profit=300units[/tex]