# C(x), of making x calculators is a quadratic function in terms of x. They have found that the cost, c(x), of making x

#1

c(x), of making x calculators is a quadratic function in terms of x. They have found that the cost, c(x), of making x calculators is a quadratic function in terms of x.

A quadratic function, which we can call c(x) like you do, in x, generally looks like this:

c(x) = ax^2 + bx + c

You see the “a”, the “b”, and the “c” in there? We don’t know those guys.
Notice that “x” means “number of calculators” and “c(x)” means “the cost of x calculators” (well, the cost of producing x calculators).

So, what we are given are various values of c(x) and x. For example, first they tell us that c(x) = 45 when x = 2, we can write that as c(2) = 45.
We also know from the problem text that c(4) = 143 and that c(10) = 869.

In the general equation we need to find 3 unknowns: a, b, and c. To find three unknowns we should have three equations. But look, they gave us three equations:

c(2) = a2^2 + b2 + c = 45 <— equation 1
c(4) = a4^2 + b4 + c = 143 <— equation 2
c(10) = a10^2 + b10 + c = 869 <— equation 3

Given those three equations we need to find a, b, and c. What do we do when we know those values? Well then if we know a, b, and c, we can get c(x) for any x (and in particular for x=1).

One way of solving such equations is by subtracting them from each other. E.g., multiply the first equation by 4 and then subtract the second equation from it:
16a + 8b + 4c = 180 <— first equation times 4
4
b + 3*c = 37 <— (equation 1 times 4) minus (equation 2)

And now we can see that:
b = (37 - 3*c)/4
So we know b in terms of c (and only c, there is no “a” in there).

Now we can plug this into equation 1, e.g., and get:
4a + 2(37 - 3c)/4 + c = 45
4
a + 18.5 - 6c/4 + c = 45
4
a - c/2 = 26.5
so:
a = (26.5 + c/2)/4
So now we know a in terms of c (and c only).
We now have b in terms of c and a in terms of c.

So let us go to the last equation, e.g., and insert a and b in terms of c. This should give us an equation entirely in terms of c. The last equation is:
100a + 10b + c = 869
now insert a and b in terms of c that we found above:
100*((26.5 + c/2)/4) + 10*((37 - 3c)/4) + c = 869
25
(26.5 + c/2) + 2.5*(37 - 3c) + c = 869
662.5 + 12.5
c + 92.5 - 7.5c + c = 869
6
c = 869 - 755
6*c = 114
c = 114/6 = 19

Okay! We’ve got c!
Remember we had this expression for b:
b = (37 - 3c)/4
we found this earlier. Since we know c now, plug in c:
b = (37 - 3
19)/4 = -5

And we had an expression for a as well:
a = (26.5 + c/2)/4
Insert c:
a = (26.5 + 19/2)/4 = 9

Okay!
So the equation is:
c(x) = 9 * x^2 - 5*x + 19

Let us verify that we have found the right values. How do we do this? Well, we know that c(2) has to be 45. So let us start there:
c(2) = 9 * 2^2 - 5*2 + 19 = 45

Yep, that works. We also know that c(4) has to be 143:
c(4) = 9 * 4^2 - 5*4 + 19 = 143

That works too! And finally, we know that c(10) must be 869:
c(10) = 9 * 10^2 - 5*10 + 19 = 869

Yay, that works as well! Okay, so now we have verified the parameters and they look really good!
The question asks us to find c(1) which is the cost of producing 1 calculator. Since we know a, b, and c, simply plug in x=1:
c(1) = 9 * 1^2 - 5*1 + 19 = 23

So we find that the cost of producing 1 calculator is 23 dollars and that is your answer. I hope this was helpful.