Given an average cost function, calculate output level

[413]

A firm faces the following Average Cost function:

AC=50q^-1 +30-1.5q+0.125q^2

Calculate the output level (q) which minimzies
a) Average Variable cost
How do i find this out if i don't know the amount of fixed cost?


b) Marginal cost
Total cost=50+30q-1.5q^2+0.125q^3
MC=30-3q+0.375q^2
0=30-3q+0.375q^2
how do you solve this? the quadratic formula doesn't seem to work

 

[Gib Z]

Basically, AC is equal to that equation. The basic theory is that if we graphed it, and found the lowest point on that graph, that would be the minimum AC.

Now, to do that by hand, we need Differential Calculus, which I hope you have done. We have the Function AC=\frac{q^2}{8} -1.5q +\frac{50}{q} +30. We take the derivative with respect to q, we get \frac{d}{dx} AC=\frac{q}{4}-1.5-\frac{50}{q^2}. Now we set that equal to zero to find the lowest point. Make some manipulations and it should work out fine.

 

[Gib Z]

O, as for b), Why doesn't the quadratic equation work? Simplified it Looks like this:
\frac{1}{8} q^2 -q-10=0
By the quadratic formula you get \frac {1+- (41)^{\frac{1}{2}}}{2}

The +- part is plus/minus, not plus the negative.

 

[dextercioby]

He's right. You messed up a "+" sign. The equation he got has indeed roots with nonzero imaginary part.

Daniel.

 

[murshid_islam]

the solution to this equation \frac{1}{8} q^2 -q-10=0 is i think 4\pm4\sqrt{6}

 

[Gib Z]

My god I am very stupid, I had a brain fart...

Edit: O wait no, i think i had it right...different equation...

 

[413]

first question:
how do i find the zeros for this equation
q/4-1.5-50/q^2
is there a way to do it by hand or you just have to graph it?

 

[Gib Z]

Hmm there is a way to do it by hand, but its a total pain in the ass And i don't think you would like it much. This is a cubic equation, which can be solved using Cardanos Method, but seriously, You will pee your pants..There are 3 zeros for this equation. If you can guess and check and find just 1, then its much much easier.

 

[413]

i used graphing calculator and found that one zero is 8.66, how can i use this and find other roots?

 

[murshid_islam]

the other roots are not real (complex). if you have drawn the curve using a graphing calculator, you can see that the curve intersects the x-axis in only one point. that point is the only real solution.

 

[413]

ok, so the only solution is 8.66

for b)
the equation i got is 0=30-3q+0.375q^2
which i have trouble finding zero for.

 

[murshid_islam]

for b) your equation is the same as
\frac{3}{8}q^2 - 3q + 30 = 0

q = \frac{3 \pm \sqrt{(-3)^2 - 4\left(\frac{3}{8}\right)(30)}}{2\left(\frac{3}{8}\right)}

but this also leads to imaginary roots.

 

[mr_homm]

I think the responses so far are somewhat too bogged down in the details of computing the solution. Take a step back and look at the original problem, which is a business/economics question. Whate value of q minimizes the marginal cost? Standard calculus procedure: determine where (a) derivative is zero, (b) derivative changes sign discontinuously, (c) the endpoints of the region of interest lie. Since this is a polynomial, (b) does not apply, and (a) leads to no solution. That leaves (c). What is the interval of interest? q represents the number of items manufactured or sold. It cannot be a negative number, so the interval of interest is the positive reals (or integers if you wish). Since the derivative is a parabola with positive leading coefficient, the derivative has a minimum but no maxumum. Since there are no zeroes, the minimum must be positive. Therefore the derivative is everywhere positive. Hence the minimum value of q is at the left endpoint of the interval of interest, i.e q=0. Conclusion, to lower your marginal cost, you should get out of this business altogether, since doing no business at all is the optimum solution.

Hope this helps!

--Stuart Anderson