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  
   $$
   R(x)=x\cdot p(x)=x\left(200-\frac{x}{30}\right)=200x-\frac{x^2}{30}.
   $$

2. This is a quadratic function in $$ x $$ with the form  
   $$
   R(x)=-\frac{1}{30}x^2+200x.
   $$
   Since the coefficient of $$ x^2 $$ is negative, the function has a maximum at the vertex.

3. The $$ x $$-coordinate of the vertex is found using  
   $$
   x=-\frac{b}{2a},
   $$
   where $$ a=-\frac{1}{30} $$ and $$ b=200 $$. Thus,
   $$
   x=-\frac{200}{2\left(-\frac{1}{30}\right)}=200\cdot\frac{30}{2}=3000.
   $$

4. Substitute $$ x=3000 $$ into $$ R(x) $$ to find the maximum revenue:
   $$
   R(3000)=200(3000)-\frac{3000^2}{30}=600\,000-\frac{9\,000\,000}{30}.
   $$
   $$
   \frac{9\,000\,000}{30}=300\,000,
   $$
   so
   $$
   R(3000)=600\,000-300\,000=300\,000.
   $$

5. Thus, the maximum revenue is  
   $$
   \$300\,000.
   $$

---

**(B) Maximum Profit**

1. The cost function is  
   $$
   C(x)=75\,000+70x.
   $$
   The profit function is defined as  
   $$
   P(x)=R(x)-C(x).
   $$
   Using the revenue function determined above, we have:
   $$
   P(x)=\left(200x-\frac{x^2}{30}\right)-\left(75\,000+70x\right).
   $$

2. Simplify the profit function:
   $$
   P(x)=200x-\frac{x^2}{30}-75\,000-70x,
   $$
   $$
   P(x)=130x-\frac{x^2}{30}-75\,000.
   $$
   This is a quadratic function in $$ x $$ of the form
   $$
   P(x)=-\frac{1}{30}x^2+130x-75\,000.
   $$

3. The maximum profit occurs at the vertex of this quadratic. The $$ x $$-coordinate is found by:
   $$
   x=-\frac{b}{2a},
   $$
   where $$ a=-\frac{1}{30} $$ and $$ b=130 $$. Accordingly,
   $$
   x=-\frac{130}{2\left(-\frac{1}{30}\right)}=130\cdot\frac{30}{2}=1950.
   $$

4. Substitute $$ x=1950 $$ into the profit function:
   $$
   P(1950)=130(1950)-\frac{1950^2}{30}-75\,000.
   $$

5. Calculate the terms:
   - $$ 130\times1950=253\,500 $$.
   - To compute $$ \frac{1950^2}{30} $$:
     $$
     1950^2=3\,802\,500 \quad \text{so} \quad \frac{3\,802\,500}{30}=126\,750.
     $$

6. Thus,
   $$
   P(1950)=253\,500-126\,750-75\,000.
   $$
   $$
   253\,500-126\,750=126\,750,
   $$
   $$
   P(1950)=126\,750-75\,000=51\,750.
   $$

7. Therefore, the maximum profit is  
   $$
   \$51\,750,
   $$
   achieved when $$ 1950 $$ television sets are manufactured and sold.

8. The price per television set is given by
   $$
   p(x)=200-\frac{x}{30}.
   $$
   Substitute $$ x=1950 $$:
   $$
   p(1950)=200-\frac{1950}{30}=200-65=135.
   $$

9. The company should charge  
   $$
   \$135 \text{ per television set}.
   $$

---

$$
\boxed{\text{(A) Maximum revenue is } \$300\,000. \ \text{(B) Maximum profit is } \$51\,750 \text{ when } 1950 \text{ sets are sold at } \$135 \text{ each.}}
$$
**(A) Maximum Revenue: \$300,000** **(B) Maximum Profit: \$51,750 when 1,950 sets are sold at \$135 each.**

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