Marginal analysis in an important topic in business calculus, and one you will very likely touch upon in your class.

In essence, marginal analysis studies how to estimate how quantities (such as profit, revenue and cost) change when the input increases by $1$.

While marginal analysis is an accurate approximation of how these quantities change when the input increases by $1$, you can also calculate the exact change, which we will cover in the sample problems.

In words: To perform marginal analysis on either profit, revenue or cost, find the derivative function for the one quantity out of these three that you are estimating for.

The derivatives of these quantities are called marginal profit function, marginal revenue function and marginal cost function, respectively.

In math notation:

If profit is given by $P(x)$, then the marginal profit function is given by $P'(x)$

If revenue is given by $R(x)$, then the marginal revenue function is given by $R'(x)$

If cost is given by $C(x)$, then the marginal cost function is given by $C'(x)$

In marginal analysis, you will usually be asked to find two things:

- an estimate of how much profit, revenue and/or cost changes when the $n^{th}$ unit is produced or sold. Here, you use derivatives.
- the exact amount of how much profit, revenue and/or cost changes. Here, you use the original function.

In other words, we can either estimate (get close to), or get the real quantity, that adding $1$ unit results in. Both approaches are explained below.

First, let's explore how to estimate changes in profit, revenue or cost.

In words: To get an estimate of how much profit, revenue and/or cost are changing for the $n^{th}$ unit, you need to find the marginal function and plug one less than $n$ (or, number of units $- \ 1$) into the marginal function

In math notation: To estimate how profit, revenue and/or cost are changing when the $n^{th}$ unit is produced or sold, plug in $n-1$ into the marginal function

For example, if you are asked to estimate how profit is changing when the $10^{th}$ unit is sold, you need to plug in $9$ (one less than $10$) into the marginal profit function.

On the other hand, calculating exactly how the quantity changes (instead of estimating) is a little different, since it uses the original function instead of the derivative.

In words: To find the exact change in profit, revenue or cost after producing or selling the $n^{th}$ unit, you need to evaluate the original function at $n$ and subtract the original function evaluated at $n-1$

In math notation: To calculate the exact change in profit, revenue or cost for the $n^{th}$ unit, calculate $f(n)-f(n-1)$, where $f(x)$ is the original function

For example, if you are asked to calculate the exact cost of producing the $14^{th}$ unit, you need to plug in both $14$ and $13$ into the original function, and subtract the latter from the former, as in $f(14)-f(13)$.

This makes sense because to find how much the $14^{th}$ unit cost, you find the cost of producing $14$ units, and subtract the cost of producing $13$ units. The remaining amount has to be what it cost to produce unit $14$.

Remember that to estimate quantities, you need to use derivatives. However, to find actual quantities, you need to use the original function instead.

Moreover, the two processes differ in which quantities you need to plug in, so make sure you understand the difference between estimating and finding actual quantities.

For a company that sells kids' toys, the total cost of producting $x$ is given by the function $$C(x)=2350+80x-0.04x^2$$ and that all $x$ toys are sold when the price is equal to $$p(x)=-2x+35$$

Question 1Estimate the marginal cost of producing the 6th unit

First, to find the marginal cost function, we simply find the derivative of the total cost function,

$$C'(x)=-0.08x+80$$

Now that we have the marginal cost function, we need to find the marginal cost of producing the $6^{th}$ unit. As explained, to estimate the change, you plug in one less than the $x$ that was given into the marginal function.

Then, since we are looking for the marginal cost of the $6^{th}$ unit, we plug in $5$ into the marginal cost function:

$$C'(5)=-0.08(5)+80$$

$$C'(5)=79.6$$

The marginal cost of producing the $6^{th}$ unit is $\$79.60$

Question 2 Calculate the actual cost of producing the 6th unit

Again, it makes sense that to find the actual cost of producing $6$ units, we calculate the cost of producing $6$ units and subtract the cost of producing $5$ units. The result of this must be the cost of producing unit $6$.

Remember: here, we are not using the marginal function anymore, since the marginal function is only used to estimate. When you are asked to find actual amounts, you will use the original profit, revenue and/or cost function.

So then, using the original cost function, we calculate,

$$C(6) - C(5)$$

$$= (-0.04(6^2)+80(6)+2350) -$$

$$(-0.04(5^2)+80(5)+2350)$$

$$=2,828.56 - 2749$$

$$C(6)-C(5)=79.56$$

So the actual cost of producing the $6^{th}$ unit is $\$79.56$. And initially, we estimated this cost would be $\$79.60$, for a difference of $4$ cents. Pretty good estimate!

Question 3 Estimate the revenue from selling the 6th unit.

Here, you need to find the marginal revenue function, which is just the derivative of the revenue function. However, we were not given a revenue function in the problem.

Fortunately, it is easy to calculuate the revenue function. Remember that revenue is simply the number of units times the price. If for example, I'm selling lemonade at $\$2$ a glass, and I sell $10$ glasses, my revenue is $10\cdot\$2=\$20$.

So the revenue function is just the number of units sold times the price of each unit.

For marginal analysis, if you don't have a revenue function but need to calculate marginal revenue, you can always get the revenue function:

$$R(x) = x \cdot p(x)$$

Which is just revenue equals to number of units times the price of each unit. After getting the revenue function, you can get the marginal revenue function by finding the derivative of the revenue function.

We proceed to calculate the revenue function,

$$R(x) = x \cdot p(x)$$

$$R(x)=(x)(-2x+35)$$

$$R(x)=-2x^2+35x$$

Now that we have the revenue function, we find the marginal revenue function (its derivative),

$$R'(x)=-4x+35$$

As before, to estimate the revenue the from selling the $6^{th}$ unit, we plug in $5$ (one less) into the marginal revenue function,

$$R'(5)=-4(5)+35$$

$$R'(5)=15$$

So, the estimated revenue of selling the $6^{th}$ unit is $\$15$.

Question 4 Calculate the actual revenue of selling the 6th unit

As we did with the cost function, we need to find the total revenue of selling the first $6$ units and subtract the revenue from selling the first $5$ units. The difference will be the revenue produced by the 6th unit,

$$R(6)-R(5)$$

$$=(-2(6^2)+35(6))- $$

$$(-2(5^2)+35(5))$$

$$=(-72+210)-(-50+175)$$

$$=138-125$$

$$=13$$

Then, the actual revenue of selling the $6^{th}$ unit is $\$13$, and our estimate was of $\$15$.

Question 5 Find the marginal profit function

To find the marginal profit function, we need to find the profit function first. However, the profit function was not given in the original problem. How is profit calculated?

Remember profit is what's left after costs are subtracted from revenues. This means that the profit function is just the revenue function minus the cost function.

For marginal analysis, we usually deal with the profit function $P(x)$, revenue function $R(x)$ and cost $C(x)$ function. If you have 2 of these quantities but need the 3rd, you can easily calculate it. Remember that profit is what you get after subtracting costs from revenue,

$$P(x) = R(x) - C(x)$$

If you need profit, we are done. But if you need either cost or revenue, just solve for it in the equation above.

In our case, we need the profit function, and we know that profit is equal to revenue minus costs:

$$P(x)=R(x)-C(x)$$

We then calculate the profit function for this specific problem,

$$P(x)=-2x^2+35x \ -$$

$$(2350 + 80x - 0.04x^2)$$

We distribute the negative sign among all terms of the cost function,

$$=-2x^2+35x-2350 - 80x + 0.04x^2$$

Finally, we combine like terms,

$$P(x)=-1.96x^2 - 45x - 2350$$

To finish, to get the marginal profit function, we find the derivative of the profit function,

$$P'(x)=3.92x-45$$

Do not confuse the profit function with the price function. The price function is usually written as $p(x)$, while the profit function is the uppercase version, $P(x)$. In summary, big $P$ is for Profit!

- Marginal analysis estimates how profit, revenue and cost change when an extra unit is produced or sold
- The marginal function of profit, revenue or cost is just its derivative function
- To estimate how a quantity is changing when the $n^{th}$ unit is produced or sold, plug in $n-1$ into the marginal function
- To calculate exactly how a quantity is changing for the $n^{th}$ unit, use the original function to find $f(n)-f(n-1)$
- Revenue is equal to number of units times price per unit, or $R(x)=x \cdot p(x)$
- Profit is equal to revenue minus costs, or $P(x)=R(x)-C(x)$. If out of profit, revenue and cost, you only have two of these quantities, use the formula just mentioned to find the third by solving for it.

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