Find the producers' surplus at a price level of $$ \bar{p}=\$ 62 $$ for the price-supply equation below. $$ p=S(x)=10+0.1 x+0.0003 x^{2} $$ The producers' surplus is $$ \$ $$. (Round to the nearest integer as needed.)

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To find the producers' surplus at a price level of $$ \bar{p} = 62 $$, we first need to determine the quantity $$ x $$ at which the supply price equals $$ \bar{p} $$. We do this by setting the supply equation equal to $$ \bar{p} $$:

$$
62 = 10 + 0.1x + 0.0003x^2
$$

Rearranging this equation gives:

$$
0.0003x^2 + 0.1x + 10 - 62 = 0
$$

$$
0.0003x^2 + 0.1x - 52 = 0
$$

Next, we can use the quadratic formula to solve for $$ x $$:

$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$

where $$ a = 0.0003 $$, $$ b = 0.1 $$, and $$ c = -52 $$.

Calculating the discriminant:

$$
b^2 - 4ac = (0.1)^2 - 4(0.0003)(-52)
$$

$$
= 0.01 + 0.0624 = 0.0724
$$

Now substituting into the quadratic formula:

$$
x = \frac{-0.1 \pm \sqrt{0.0724}}{2 \cdot 0.0003}
$$

Calculating $$ \sqrt{0.0724} $$:

$$
\sqrt{0.0724} = 0.26
$$

Now substituting back:

$$
x = \frac{-0.1 \pm 0.26}{0.0006}
$$

Calculating the two possible values for $$ x $$:

1. Using the positive root:

$$
x = \frac{0.16}{0.0006} \approx 266.67
$$

2. Using the negative root (not applicable since quantity cannot be negative):

$$
x = \frac{-0.36}{0.0006} \approx -600 \quad (\text{not valid})
$$

Thus, we take $$ x \approx 266.67 $$. We round this to $$ x = 267 $$ for practical purposes.

Next, we calculate the producers' surplus, which is given by the area above the supply curve and below the price level $$ \bar{p} $$ from $$ x = 0 $$ to $$ x = 267 $$.

The producers' surplus is calculated as:

$$
\text{Producers' Surplus} = \int_0^{267} \bar{p} \, dx - \int_0^{267} S(x) \, dx
$$

Calculating the first integral:

$$
\int_0^{267} 62 \, dx = 62 \cdot 267 = 16554
$$

Now we calculate the second integral:

$$
\int_0^{267} S(x) \, dx = \int_0^{267} (10 + 0.1x + 0.0003x^2) \, dx
$$

Calculating each term separately:

1. For $$ 10 $$:

$$
\int_0^{267} 10 \, dx = 10 \cdot 267 = 2670
$$

2. For $$ 0.1x $$:

$$
\int_0^{267} 0.1x \, dx = 0.1 \cdot \frac{x^2}{2} \bigg|_0^{267} = 0.1 \cdot \frac{267^2}{2} = 0.1 \cdot \frac{71289}{2} = 0.1 \cdot 35644.5 = 3564.45
$$

3. For $$ 0.0003x^2 $$:

$$
\int_0^{267} 0.0003x^2 \, dx = 0.0003 \cdot \frac{x^3}{3} \bigg|_0^{267} = 0.0003 \cdot \frac{267^3}{3} = 0.0003 \cdot \frac{19066263}{3} = 0.0003 \cdot 6355421 = 1906.6263
$$

Now summing these results:

$$
\int_0^{267} S(x) \, dx = 2670 + 3564.45 + 1906.6263 \approx 8141.0763
$$

Finally, we calculate the producers' surplus:

$$
\text{Producers' Surplus} = 16554 - 8141.0763 \approx 8402.9237
$$

Rounding to the nearest integer, the producers' surplus is:

$$
\boxed{8403}
$$
The producers' surplus at a price level of $$ \$62 $$ is approximately $$ \$8403 $$.

Find the producers’ surplus at a price level of for the price-supply equation below.

The producers’ surplus is .
(Round to the nearest integer as needed.)

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Find the producers’ surplus at a price level of for the price-supply equation below. The producers’ surplus is . (Round to the nearest integer as needed.)