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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**(A) Maximum Revenue**

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

2. This is a quadratic function in $$ x $$ with a negative coefficient for $$ x^2 $$ (i.e. $$ -\frac{1}{30} $$), so its graph is a downward-opening parabola. The maximum revenue occurs at the vertex.

3. The vertex for a quadratic $$ R(x) = ax^2 + bx + c $$ is at:
   $$
   x = -\frac{b}{2a}
   $$
   Here, $$ a = -\frac{1}{30} $$ and $$ b = 200 $$. Therefore:
   $$
   x = -\frac{200}{2\left(-\frac{1}{30}\right)} = \frac{200}{\frac{2}{30}} = 200 \times \frac{30}{2} = 3000
   $$

4. Substitute $$ x = 3000 $$ into $$ R(x) $$:
   $$
   R(3000) = 200(3000) - \frac{3000^2}{30} = 600000 - \frac{9000000}{30} = 600000 - 300000 = 300000
   $$

Thus, the maximum revenue is $$\$300000$$.

**(B) Maximum Profit**

1. The cost function is:
   $$
   C(x) = 75000 + 70x
   $$
   and the revenue function is:
   $$
   R(x) = 200x - \frac{x^2}{30}
   $$
   The profit function is:
   $$
   P(x) = R(x) - C(x) = \left(200x - \frac{x^2}{30}\right) - (75000 + 70x)
   $$

2. Simplify the profit function:
   $$
   P(x) = 200x - \frac{x^2}{30} - 75000 - 70x = 130x - 75000 - \frac{x^2}{30}
   $$
   or rewritten as:
   $$
   P(x) = -\frac{1}{30}x^2 + 130x - 75000
   $$

3. This is again a quadratic function in $$ x $$ with $$ a = -\frac{1}{30} $$ and $$ b = 130 $$. Its maximum occurs at:
   $$
   x = -\frac{b}{2a} = -\frac{130}{2\left(-\frac{1}{30}\right)} = \frac{130}{\frac{2}{30}} = 130 \times \frac{30}{2} = 1950
   $$

4. Substitute $$ x = 1950 $$ into $$ P(x) $$ to determine the maximum profit:
   $$
   P(1950) = -\frac{1}{30}(1950)^2 + 130(1950) - 75000
   $$
   
   First, calculate $$ 130 \times 1950 $$:
   $$
   130 \times 1950 = 253500
   $$
   
   Next, compute $$ (1950)^2 $$:
   $$
   1950^2 = 3802500
   $$
   
   Then, divide by 30:
   $$
   \frac{3802500}{30} = 126750
   $$
   
   Now, substitute back:
   $$
   P(1950) = -126750 + 253500 - 75000 = 253500 - 201750 = 51750
   $$
   
   Thus, the maximum profit is $$\$51750$$.

5. To find the price per television set when $$ x = 1950 $$, use the demand equation:
   $$
   p(x) = 200 - \frac{x}{30}
   $$
   Substitute $$ x = 1950 $$:
   $$
   p(1950) = 200 - \frac{1950}{30} = 200 - 65 = 135
   $$

Therefore, the company should charge $$\$135$$ for each television set.

**Final Answers:**

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

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