Question

For a certain company , the cost for producing items is 45x + 300 and the revenue for selling items is 85x - 0.5x ^ 2. Part a: set up an expression for the profit from producing and selling x items and solve. we assume the company sells all of the items it produces. Part B: find two values of x thatvwill create a profit of $50. Part C: is it possible for the company to make a profit if $2500?

Answer 1

Producing cost:

$45x+300$

Revenue:

$85x-0.5x^2$

The profit (P) is equal to substract the producing cost for the revenue:

$\begin{gathered} P=(85x-0.5x^2)-(45x+300) \\ P=85x-0.5x^2-45x-300 \\ P=40x-0.5x^2-300 \end{gathered}$

You can write also as:

$P=-0.5x^2+40x-300$

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P=50:

$50=-0.5x^2+40x-300$

To solve for x:

Substract 50 in both sides of the equation:

$\begin{gathered} 50-50=-0.5x^2+40x-300-50 \\ 0=-0.5x^2+40x-350 \end{gathered}$

Use the quadratic formula:

$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$$\begin{gathered} x=\frac{-40\pm\sqrt[]{40^2-4(-0.5)(-350)}}{2(-0.5)} \\ \\ x=\frac{-40\pm\sqrt[]{1600-700}}{-1} \\ \\ x=\frac{-40\pm\sqrt[]{900}}{-1} \\ \\ x=\frac{-40\pm30}{-1} \\ \\ x_1=\frac{-40+30}{-1}=\frac{-10}{-1}=10 \\ \\ x_2=\frac{-40-30}{-1}=\frac{-70}{-1}=70 \end{gathered}$Then, the two values of x that make a profit of $50 are: 10 and 70

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P=2500

$2500=-0.5x^2+40x-300$

Solve for x:

$\begin{gathered} 0=-0.5x^2+40x-300-2500 \\ 0=-0.5x^2+40x-2800 \\ \\ x=\frac{-40\pm\sqrt[]{40^2-4(-0.5)(-2500)}}{2(-0.5)} \\ \\ x=\frac{-40\pm\sqrt[]{1600-5000}}{-1} \\ \\ x=\frac{-40\pm\sqrt[]{-34000}}{-1} \end{gathered}$

As the number under the square root is a negative number the equation has no solution (value of x) in the real numbers.

No, is not possible for the company to make a profit of $2500