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

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

**Step 2.**  
To find the maximum revenue, we need to determine the critical point by differentiating $$ R(x) $$ and setting the derivative equal to zero. The derivative is  
$$
R'(x)=\frac{d}{dx}\left(200x-\frac{x^2}{30}\right)=200-\frac{2x}{30}=200-\frac{x}{15}.
$$

Setting $$ R'(x)=0 $$, we solve  
$$
200-\frac{x}{15}=0 \quad\Longrightarrow\quad \frac{x}{15}=200 \quad\Longrightarrow\quad x=200\cdot 15=3000.
$$

**Step 3.**  
Substitute $$ x=3000 $$ into the revenue function:  
$$
R(3000)=200(3000)-\frac{3000^2}{30}=600000-\frac{9000000}{30}=600000-300000=300000.
$$

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

---

**Step 4.**  
The cost function is given by  
$$
C(x)=75000+70x.
$$

The profit function $$ P(x) $$ is defined as revenue minus cost:  
$$
P(x)=R(x)-C(x)=\left(200x-\frac{x^2}{30}\right)-\left(75000+70x\right).
$$
Simplify the profit function:  
$$
P(x)=200x-\frac{x^2}{30}-75000-70x=130x-\frac{x^2}{30}-75000.
$$

**Step 5.**  
To maximize profit, differentiate $$ P(x) $$ with respect to $$ x $$:  
$$
P'(x)=\frac{d}{dx}\left(130x-\frac{x^2}{30}-75000\right)=130-\frac{2x}{30}=130-\frac{x}{15}.
$$

Setting $$ P'(x)=0 $$:  
$$
130-\frac{x}{15}=0 \quad\Longrightarrow\quad \frac{x}{15}=130 \quad\Longrightarrow\quad x=130\cdot 15=1950.
$$

**Step 6.**  
Substitute $$ x=1950 $$ into the profit function:  
$$
P(1950)=130(1950)-\frac{1950^2}{30}-75000.
$$

First, calculate  
$$
130(1950)=253500.
$$

Next, compute $$ 1950^2 $$:  
$$
1950^2=1950\cdot1950=3802500.
$$

Then,  
$$
\frac{3802500}{30}=126750.
$$

Finally,  
$$
P(1950)=253500-126750-75000=253500-201750=51750.
$$

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

**Step 7.**  
Determine the price at the production level $$ x=1950 $$:  
$$
p(x)=200-\frac{x}{30}.
$$

So,  
$$
p(1950)=200-\frac{1950}{30}=200-65=135.
$$

---

**Final Answers:**  
(A) The maximum revenue is $$\$300000$$.  
(B) The maximum profit is $$\$51750$$ when $$1950$$ sets are manufactured and sold at $$\$135$$ each.  
$$
\boxed{\text{(A) } 300000, \quad \text{(B) } 51750,\ 1950,\ 135.}
$$
(A) The maximum revenue is \$300,000. (B) The maximum profit is \$51,750 when 1,950 sets are manufactured and sold at \$135 each.

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