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=(-40\pm30)/(-1) \\ \\ x_1=(-40+30)/(-1)=(-10)/(-1)=10 \\ \\ x_2=(-40-30)/(-1)=(-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