mcr3u unit 1 test due: september 24th @10:30am app /18 t&i /23 8. the cost function for a telephone…

mcr3u unit 1 test due: september 24th @10:30am app /18 t&i /23 8. the cost function for a telephone manufacturing plant is c(x)=5x + 24 and the revenue function is r(x)=-2x^{2}+27x, where x is the number of telephones sold, in thousands. a) determine the profit function for the number of phones sold. 2t b) determine the number of telephones sold that maximizes profit the maximum profit in thousands. 5t

mcr3u unit 1 test due: september 24th @10:30am app /18 t&i /23 8. the cost function for a telephone manufacturing plant is c(x)=5x + 24 and the revenue function is r(x)=-2x^{2}+27x, where x is the number of telephones sold, in thousands. a) determine the profit function for the number of phones sold. 2t b) determine the number of telephones sold that maximizes profit the maximum profit in thousands. 5t

Answer

Explanation:

Step1: Recall profit - revenue - cost formula

The profit function $P(x)$ is given by $P(x)=R(x)-C(x)$. Given $C(x) = 5x + 24$ and $R(x)=-2x^{2}+27x$.

Step2: Substitute the functions

$P(x)=(-2x^{2}+27x)-(5x + 24)$. $P(x)=-2x^{2}+27x - 5x-24$. $P(x)=-2x^{2}+22x - 24$.

Step3: For part b, find the vertex of the quadratic function

The quadratic function $P(x)=-2x^{2}+22x - 24$ is of the form $y = ax^{2}+bx + c$, where $a=-2$, $b = 22$, and $c=-24$. The $x$ - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. $x=-\frac{22}{2\times(-2)}=\frac{22}{4}=5.5$.

Step4: Find the maximum profit

Substitute $x = 5.5$ into the profit function $P(x)$. $P(5.5)=-2\times(5.5)^{2}+22\times5.5 - 24$. $P(5.5)=-2\times30.25+121 - 24$. $P(5.5)=-60.5+121 - 24$. $P(5.5)=36.5$.

Answer:

a) $P(x)=-2x^{2}+22x - 24$ b) The number of telephones sold (in thousands) that maximizes profit is $x = 5.5$, and the maximum profit (in thousands) is $36.5$.