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**

The revenue function is given by
$$
R(x) = x \cdot p(x) = x\left(200 - \frac{x}{30}\right) = 200x - \frac{x^2}{30}.
$$
This is a quadratic function that opens downward, so its maximum occurs at the vertex. The $$x$$-coordinate of the vertex is found by
$$
x = -\frac{b}{2a}.
$$
Here, writing the equation in the standard form $$R(x) = ax^2 + bx + c$$, we have
$$
a = -\frac{1}{30} \quad \text{and} \quad b = 200.
$$
Thus,
$$
x = -\frac{200}{2\left(-\frac{1}{30}\right)} = \frac{200}{\frac{2}{30}} = 200 \times \frac{30}{2} = 3000.
$$
The maximum revenue is then
$$
R(3000) = 200(3000) - \frac{3000^2}{30} = 600000 - \frac{9000000}{30} = 600000 - 300000 = 300000.
$$

**(A) Answer:** The maximum revenue is $$ \$300000 $$.

---

**(B) Maximum Profit**

The profit function is
$$
P(x) = R(x) - C(x),
$$
where the cost function is
$$
C(x) = 75000 + 70x.
$$
Substitute the revenue function:
$$
P(x) = \left(200x - \frac{x^2}{30}\right) - (75000 + 70x) = 200x - \frac{x^2}{30} - 75000 - 70x.
$$
Simplify:
$$
P(x) = 130x - \frac{x^2}{30} - 75000.
$$
This is also a quadratic function (opening downward) and reaches its maximum at the vertex:
$$
x = -\frac{b}{2a}.
$$
In this expression, treating it as
$$
P(x) = -\frac{1}{30}x^2 + 130x - 75000,
$$
we have
$$
a = -\frac{1}{30} \quad \text{and} \quad b = 130.
$$
Thus,
$$
x = -\frac{130}{2\left(-\frac{1}{30}\right)} = \frac{130}{\frac{2}{30}} = 130 \times \frac{30}{2} = 1950.
$$
Now calculate the maximum profit:
$$
P(1950) = 130(1950) - \frac{1950^2}{30} - 75000.
$$
Compute:
$$
130 \times 1950 = 253500.
$$
$$
1950^2 = 3802500, \quad \frac{3802500}{30} = 126750.
$$
Thus,
$$
P(1950) = 253500 - 126750 - 75000 = 253500 - 201750 = 51750.
$$

The price at $$ x = 1950 $$ is:
$$
p(1950) = 200 - \frac{1950}{30} = 200 - 65 = 135.
$$

**(B) Answer:** 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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