b) determine the number of telephones sold that maximizes profit and the maximum profit in thousands. 5t

Explanation:

Step1: Identify the profit - function

Let the profit function $P(x)=-2x^{2}+27x - 5x-24=-2x^{2}+22x - 24$, where $x$ is the number of telephones sold (in thousands).

Step2: Find the derivative

The derivative of $P(x)$ using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ is $P^\prime(x)=\frac{d}{dx}(-2x^{2}+22x - 24)=-4x + 22$.

Step3: Set the derivative equal to zero

To find the critical points, set $P^\prime(x)=0$. So, $-4x + 22 = 0$. Solve for $x$: [ \begin{align*} -4x&=-22\ x&=\frac{22}{4}=\frac{11}{2}=5.5 \end{align*} ]

Step4: Find the second - derivative

The second - derivative $P^{\prime\prime}(x)=\frac{d}{dx}(-4x + 22)=-4$. Since $P^{\prime\prime}(x)=-4<0$, the function $P(x)$ has a maximum at $x = 5.5$.

Step5: Find the maximum profit

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

Answer:

The number of telephones sold to maximize profit is $5.5$ thousand, and the maximum profit is $36.5$ thousand.