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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Alright, I need to solve this problem involving a company that manufactures and sells $$ x $$ television sets per month. The given equations are the cost function $$ C(x) = 75,000 + 70x $$ and the price-demand equation $$ p(x) = 200 - \frac{x}{30} $$, with $$ 0 \leq x \leq 6000 $$. The questions are:

(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.

First, let's recall some fundamental concepts:

1. **Revenue ($$ R(x) $$)**: This is the total income from selling $$ x $$ units. It's calculated as the product of the number of units sold and the price per unit.
   
   $$
   R(x) = x \times p(x)
   $$
   
2. **Profit ($$ P(x) $$)**: This is the difference between the revenue and the total cost.
   
   $$
   P(x) = R(x) - C(x)
   $$
   
Given the equations:
- $$ C(x) = 75,000 + 70x $$
- $$ p(x) = 200 - \frac{x}{30} $$

**Part (A): Maximum Revenue**

To find the maximum revenue, I need to determine the value of $$ x $$ that maximizes $$ R(x) $$. Since $$ R(x) = x \times p(x) $$, and $$ p(x) $$ is a linear function, $$ R(x) $$ will also be a quadratic function.

Let's express $$ R(x) $$ in terms of $$ x $$:
$$
R(x) = x \times \left(200 - \frac{x}{30}\right) = 200x - \frac{x^2}{30}
$$

This is a quadratic equation in the form $$ R(x) = -\frac{1}{30}x^2 + 200x $$. The coefficient of $$ x^2 $$ is negative, which means the parabola opens downward, and the vertex of this parabola will give the maximum revenue.

The vertex of a parabola $$ ax^2 + bx + c $$ is at $$ x = -\frac{b}{2a} $$. Here, $$ a = -\frac{1}{30} $$ and $$ b = 200 $$.

$$
x = -\frac{200}{2 \times \left(-\frac{1}{30}\right)} = -\frac{200}{-\frac{2}{30}} = 200 \times \frac{30}{2} = 3000
$$

So, the maximum revenue occurs when $$ x = 3000 $$ television sets are sold.

Now, let's calculate the maximum revenue:
$$
R(3000) = 3000 \times \left(200 - \frac{3000}{30}\right) = 3000 \times (200 - 100) = 3000 \times 100 = \$300,000
$$

**Part (B): Maximum Profit**

To find the maximum profit, I need to determine the value of $$ x $$ that maximizes $$ P(x) $$. Let's express $$ P(x) $$ in terms of $$ x $$:
$$
P(x) = R(x) - C(x) = \left(200x - \frac{x^2}{30}\right) - (75,000 + 70x) = -\frac{x^2}{30} + 130x - 75,000
$$

This is another quadratic equation in the form $$ P(x) = -\frac{1}{30}x^2 + 130x - 75,000 $$. Again, the coefficient of $$ x^2 $$ is negative, so the parabola opens downward, and the vertex will give the maximum profit.

Using the vertex formula:
$$
x = -\frac{130}{2 \times \left(-\frac{1}{30}\right)} = -\frac{130}{-\frac{2}{30}} = 130 \times \frac{30}{2} = 1950
$$

So, the maximum profit occurs when $$ x = 1950 $$ television sets are sold.

Now, let's calculate the maximum profit:
$$
P(1950) = -\frac{(1950)^2}{30} + 130 \times 1950 - 75,000
$$
$$
= -\frac{3,802,500}{30} + 253,500 - 75,000
$$
$$
= -126,750 + 253,500 - 75,000
$$
$$
= 51,750
$$

**Price per Television Set:**
To find the price the company should charge for each television set when producing 1950 units, we use the price-demand equation:
$$
p(1950) = 200 - \frac{1950}{30} = 200 - 65 = \$135
$$

**Summary:**
- **Maximum Revenue**: \$300,000 when 3000 television sets are sold.
- **Maximum Profit**: \$51,750 when 1950 television sets are manufactured and sold at \$135 each.

(A) The maximum revenue is \$300,000. (B) The maximum profit is \$51,750 when 1950 sets are manufactured and sold for \$135 each.

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