A company manufactures and sells $$ x $$ television sets per month. The monthly cost and price-demand equations are $$ C(x)=75,000+70 x $$ and $$ p(x)=200-\frac{x}{30}, 0 \leq x \leq 6000 $$. (A) Find the maximum revenue. (B) Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. (A) The maximum revenue is $$ \$ \square $$ (Type an integer or a decimal.) (B) The maximum profit is $$ \$ \square $$ when $$ \square $$ sets are manufactured and sold for $$ \$ \square $$ each. (Type integers or decimals.)

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UpStudy AI · Accepted Answer

**(A) Maximum Revenue**

The revenue function is given by  
$$
R(x) = x \, p(x) = x\left(200-\frac{x}{30}\right) = 200x - \frac{x^2}{30}.
$$

This is a quadratic function that opens downward. To find its maximum, we compute the derivative and set it equal to zero:
$$
R'(x) = 200 - \frac{2x}{30} = 200 - \frac{x}{15}.
$$
Setting the derivative equal to zero:
$$
200 - \frac{x}{15} = 0 \quad \Longrightarrow \quad \frac{x}{15} = 200 \quad \Longrightarrow \quad x = 200 \times 15 = 3000.
$$

Plug $$ x = 3000 $$ back into the revenue function:
$$
R(3000) = 200(3000) - \frac{3000^2}{30} = 600\,000 - \frac{9\,000\,000}{30} = 600\,000 - 300\,000 = 300\,000.
$$

Thus, the maximum revenue is  
$$
\boxed{300000}.
$$

---

**(B) Maximum Profit**

The profit function is defined as revenue minus cost:
$$
P(x) = R(x) - C(x).
$$
Given the cost function  
$$
C(x)=75\,000 + 70x,
$$
and the revenue function we already have, the profit function becomes:
$$
P(x) = \left(200x - \frac{x^2}{30}\right) - \left(75\,000 + 70x\right) = 200x - \frac{x^2}{30} - 75\,000 - 70x.
$$
Simplify:
$$
P(x) = 130x - \frac{x^2}{30} - 75\,000.
$$

This quadratic function is also downward opening. We find the maximum profit by taking the derivative and setting it equal to zero:
$$
P'(x) = 130 - \frac{2x}{30} = 130 - \frac{x}{15}.
$$
Set $$ P'(x)=0 $$:
$$
130 - \frac{x}{15} = 0 \quad \Longrightarrow \quad \frac{x}{15} = 130 \quad \Longrightarrow \quad x = 130 \times 15 = 1950.
$$

Now, calculate the maximum profit $$ P(1950) $$:
$$
P(1950) = 130(1950) - \frac{1950^2}{30} - 75\,000.
$$

First, compute $$ 130 \times 1950 $$:
$$
130 \times 1950 = 253\,500.
$$

Next, compute $$ \frac{1950^2}{30} $$:
$$
1950^2 = 1950 \times 1950 = 3\,802\,500,
$$
$$
\frac{3\,802\,500}{30} = 126\,750.
$$

Now, substitute these into $$ P(1950) $$:
$$
P(1950) = 253\,500 - 126\,750 - 75\,000 = 51\,750.
$$

Thus, the maximum profit is  
$$
\boxed{51750}.
$$

The price at which the televisions should be sold is found using the price-demand equation:
$$
p(x)=200-\frac{x}{30}.
$$
Substitute $$ x = 1950 $$:
$$
p(1950) = 200 - \frac{1950}{30} = 200 - 65 = 135.
$$

So, the company should charge  
$$
\boxed{135}
$$
dollars per television set.

---

**Final Answers:**  
(A) The maximum revenue is $$ \$300000 $$.  
(B) The maximum profit is $$ \$51750 $$ when $$ 1950 $$ sets are manufactured and sold for $$ \$135 $$ each.
**(A) Maximum Revenue:** \$300,000 **(B) Maximum Profit:** \$51,750 when 1,950 sets are manufactured and sold for \$135 each.

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