Question

suppose the revenue from producing (and selling) x units of a product is given by ( r(x) = 9x - 0.02x^2 ) dollars. (a) find the marginal revenue at a production level of 30. (b) find the production levels where the revenue is $700. (a) the marginal revenue at a production level of 30 is 7.8 dollars per unit. (b) the production levels where the revenue is $700 are \\(\square\\). (use a comma to separate answers as needed.)

Explanation:

Step1: Set up the revenue equation

We know that the revenue function is \( R(x) = 9x - 0.02x^2 \). We want to find the production levels \( x \) where \( R(x)=700 \). So we set up the equation:
\( 9x - 0.02x^2 = 700 \)

Step2: Rearrange the equation to standard quadratic form

First, multiply both sides of the equation by 100 to eliminate the decimal:
\( 900x - 2x^2 = 70000 \)
Then, rearrange the terms to get a standard quadratic equation \( ax^2+bx + c = 0 \):
\( 2x^2 - 900x + 70000 = 0 \)
We can simplify this equation by dividing all terms by 2:
\( x^2 - 450x + 35000 = 0 \)

Step3: Solve the quadratic equation

We can solve the quadratic equation \( x^2 - 450x + 35000 = 0 \) using the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b=- 450 \), and \( c = 35000 \).

First, calculate the discriminant \( \Delta=b^2-4ac \):
\( \Delta=(-450)^2-4\times1\times35000 \)
\( \Delta = 202500-140000 \)
\( \Delta = 62500 \)

Then, find the square root of the discriminant: \( \sqrt{\Delta}=\sqrt{62500} = 250 \)

Now, substitute into the quadratic formula:
\( x=\frac{450\pm250}{2} \)

We have two solutions:

• For the plus sign: \( x_1=\frac{450 + 250}{2}=\frac{700}{2}=350 \)
• For the minus sign: \( x_2=\frac{450-250}{2}=\frac{200}{2} = 100 \)

Answer:

100, 350