# How to adapt a profit function when given new variable cost

For a cost function I have been given the following information:

fixed costs: 65\$Production of 10 elements costs: 80\$
Production of 20 elements costs: 87\$Producing 1 more when producing 10 elements costs 1\$

[Edit:] And I know that it has to be a cubic function.

And have arrived at the following cost function:

$C_1(x)=0.001x^3-0.07x^2+2.1x+65$

[Edit:] (Which I know is correct, because it is given as a control point.)

I have the revenue function:

$R(x)=8x$

Which leads to the profit function:

$P_1(x)=-0.001x^3+0.07x^2+5.9x-65$

Now another product should be produced where

variable cost per element is higher by 1.1\$fixed costs for the new product is 20\$

The profit function for the new product is given as:

$P_2(x)=-0.001x^3+0.07x^2+5.8x-20$

All that is left for me to do is prove that this is correct.

Any idea how to do this?

Thank you!

X/Y SPOILER

I have calculated variable costs for the original product:

$cv_1(10) = 15$
$cv_1(20) = 22$

so I got variable costs for the second product:

$cv_2(10) = 26$
$cv_2(20) = 44$

And have tried to adapt the original system of equations with these values, and solve them, but I am not arriving at the correct equatation.

Let $C_1(x)$ be the cost function you're looking for, which takes the number of elements produced, $x$, and returns the cost in dollars to produce that many elements. Each of the four pieces of information you're given can be interpreted as mathematical statements.

• We know the fucntion is a cubic, which means there exist $a,b,c,d \in \mathbb{R}$ so that $$C_1(x) = ax^3 + bx^2 + cx + d$$

• The fixed costs are $\$65$. This means $$C_1(0) = 65 \qquad\!\!\! \Rightarrow \qquad\!\!\!d=65$$ • Production of ten elements costs$\$80$. This means $$C_1(10) = 80 \qquad\!\!\! \Rightarrow \qquad\!\!\! 1000a + 100b + 10c = 15$$

• Production of twenty elements costs $\$87$. This means $$C_1(20) = 87 \qquad\!\!\! \Rightarrow \qquad\!\!\! 8000a + 400b + 20c = 22$$ • Producing one more when producing ten elements costs$\$1$. This means $$C_1'(10) =1$$

The derivative of $C_1(x)$ is $$C_1'(x) = 3ax^2 + 2bx + c$$

at $x=10$, this gives

$$1 = C_1'(10) = 300a + 20b + c$$

Now we have three linear equations with three variables $a,b$, and $c$.

$$1000a + 100b + 10c = 15$$

$$3000a + 200b + 10c = 10$$

$$8000a + 400b + 20c = 22$$

Some straightforward cancellation gives the same values you found. We have $$C_1(x) = 0.001x^3 -0.07x^2 + 2.1x + 65$$

Given this and the revenue function $R(x) = 8x$, your calculation of the profit function $P_1(x)$ using $R(x) = P_1(x) - C_1(x)$ is correct.

Now we have a modified cost function, $C_2(x)$, in which the variable costs per element are higher by $\$1.1$. This means $$C_2'(x)-C_1'(x) = 1.1$$ This means that the third coefficient ($c$) in the cost function is higher by$1.1$. We also know that the fixed costs are$\$20$ instead of the previous $\$65$. This means that the fourth coefficient ($d$) in the cost function is now$\$20$.

We can now write the modified cost function: $$C_2(x) = 0.001x^3 -0.07x^2+3.2x + 20$$

Given the same $R(x) = 8x$, you should now be able to derive $P_2(x)$ using $R(x) = P_2(x) - C_2(x)$.

• Added some clarification to thr original question. – ingo Nov 29 '14 at 19:15
• Okay, makes a big difference that it's cubic :) – Zubin Mukerjee Nov 29 '14 at 19:17
• I've edited the answer. Hope it helps. I really only get to answering your question in the third part of it; the first two are verifying what you've already done. – Zubin Mukerjee Nov 29 '14 at 19:35
• Hi Zubin,thank you. I added a new answer, because I wanted to use formatting to ask for some clarification. I'm not sure, wether this was the right thing to do. – ingo Nov 29 '14 at 20:35

$C'(x)$ computes the increase of $C(x)$.
At every $x$ the increase is \$1.1 higher, thus$C_2'(x)$values must be 1.1 than$C_1'(x)$for every$x$. When differentiating c becomes the absolute member of$C'(x)\$.