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

**(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 in $$ x $$ that opens downward, so its maximum occurs at the vertex. The $$ x $$-coordinate of the vertex for a quadratic $$ ax^2+bx+c $$ is  
$$
x=-\frac{b}{2a}.
$$
Here,  
$$
a=-\frac{1}{30} \quad \text{and} \quad b=200.
$$
Thus,  
$$
x=-\frac{200}{2\left(-\frac{1}{30}\right)}=200\cdot\frac{30}{2}=3000.
$$

Now, substitute $$ x=3000 $$ into the revenue function:  
$$
R(3000)=200(3000)-\frac{(3000)^2}{30}.
$$

First, compute each term:  
$$
200(3000)=600000, \quad \text{and} \quad \frac{3000^2}{30}=\frac{9\,000\,000}{30}=300000.
$$

Thus,  
$$
R(3000)=600000-300000=300000.
$$

The maximum revenue is  
$$
\$300000.
$$

---

**(B) Maximum Profit**

The profit function is given by  
$$
P(x)=R(x)-C(x).
$$
We already have  
$$
R(x)=200x-\frac{x^2}{30},
$$
and the cost function is  
$$
C(x)=75000+70x.
$$
Thus,  
$$
P(x)=\left(200x-\frac{x^2}{30}\right)-\left(75000+70x\right)=130x-\frac{x^2}{30}-75000.
$$

This quadratic function (in $$ x $$) has a maximum at its vertex. Its vertex is at  
$$
x=-\frac{b}{2a},
$$
where now the quadratic is written in the form  
$$
-\frac{1}{30}x^2+130x-75000.
$$
Here,  
$$
a=-\frac{1}{30} \quad \text{and} \quad b=130.
$$
Thus,  
$$
x=-\frac{130}{2\left(-\frac{1}{30}\right)}=130\cdot\frac{30}{2}=1950.
$$

Now compute the maximum profit $$ P(1950) $$:  
$$
P(1950)=130(1950)-\frac{(1950)^2}{30}-75000.
$$

Calculate the components:

1. $$ 130(1950)=253500 $$.

2. For $$\frac{(1950)^2}{30}$$:  
   $$
   1950^2=3802500,
   $$
   so  
   $$
   \frac{3802500}{30}=126750.
   $$

Thus,  
$$
P(1950)=253500-126750-75000.
$$
Compute the subtraction:  
$$
253500-126750=126750,
$$
and then  
$$
126750-75000=51750.
$$

The maximum profit is  
$$
\$51750.
$$

The price to charge when $$ x=1950 $$ is found from the price-demand equation:  
$$
p(1950)=200-\frac{1950}{30}.
$$
Calculate:  
$$
\frac{1950}{30}=65,
$$
so  
$$
p(1950)=200-65=135.
$$

Therefore, the maximum profit of $$\$51750$$ is realized when $$1950$$ sets are manufactured and sold at a price of $$\$135$$ each.

---

**Final Answers:**  
(A) The maximum revenue is $$\$300000$$.  
(B) 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 at \$135 each.

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