The marginal cost (MC) and marginal revenue (MR) functions of a product are $MC = 20 + \frac{x}{20}$ and $MR = 30$ respectively. If the fixed cost is 200, then the maximum value of the profit is:

Understanding the Problem:

Marginal Cost (MC): The cost to produce one additional unit

Marginal Revenue (MR): The revenue earned from selling one additional unit

Fixed Cost: Costs that don't change regardless of production (like rent) $= 200$

Profit: Total Revenue - Total Cost

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To get total functions from marginal functions, we integrate:

Total Cost (TC):

$TC = \int MC \, dx = \int \left(20 + \frac{x}{20}\right) dx$

$TC = 20x + \frac{x^2}{40} + C$

Since fixed cost $= 200$, when $x = 0$, $TC = 200$, so $C = 200$

$TC = 20x + \frac{x^2}{40} + 200$

Total Revenue (TR):

$TR = \int MR \, dx = \int 30 \, dx$

$TR = 30x$

(No constant needed since revenue is $0$ when no units are sold)

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Profit Function:

$\text{Profit} = TR - TC$

$P(x) = 30x - \left(20x + \frac{x^2}{40} + 200\right)$

$P(x) = 30x - 20x - \frac{x^2}{40} - 200$

$P(x) = 10x - \frac{x^2}{40} - 200$

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To maximize profit, set the derivative equal to zero:

$\frac{dP}{dx} = 10 - \frac{2x}{40} = 10 - \frac{x}{20}$

Setting to zero:

$10 - \frac{x}{20} = 0$

$10 = \frac{x}{20}$

$x = 200$

Verifying it's a maximum:

$\frac{d^2P}{dx^2} = -\frac{1}{20} < 0$

Negative second derivative confirms this is a maximum.

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Substitute $x = 200$ into profit function:

$P(200) = 10(200) - \frac{(200)^2}{40} - 200$

$= 2000 - \frac{40000}{40} - 200$

$= 2000 - 1000 - 200$

$= 800$

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Maximum Profit $= 800$