Okay, I have a problem that I have been thinking of for some time now.
Marginal Cost as everyone knows, crosses the average total cost (ATC) and the average variable cost (AVC) at each of their minimums. The thing is, how is that mathematically proved?
I can think of a potential proof given that the total cost (TC) can be denotated as a polynomial function:
ax^n+bx^(n-1)+....+cx+d (TC)
Marginal cost is just the derivative so it would be:
anx^(n-1)+b(n-1)x^(n-2)+....+c (MC)
Average total cost is the total cost divided by the quantity, which would be x in this case so ATC would be
ax^(n-1)+bx^(n-2)+....+c+d/x (ATC)
If you set MC=ATC,
anx^(n-1)+b(n-1)x^(n-2)+....+c = ax^(n-1)+bx^(n-2)+....+c+d/x
multiply by x on both sides
anx^n+b(n-1)x^(n-1)+....+cx = ax^n+bx^(n-1)+....+cx+d
And use simple algebra, you get
a(n-1)x^n+b(n-2)x^(n-1)+....= d
which is essentially unsolveable
However, if you take the derivative of the ATC and solve for 0 to find the minimum,
a(n-1)x^(n-2)+b(n-2)x^(n-3)+....-d/(x^2) (derivative of ATC)
a(n-1)x^(n-2)+b(n-2)x^(n-3)+....-d/(x^2)=0
add by d/(x^2)
a(n-1)x^(n-2)+b(n-2)x^(n-3)+....=d/(x^2)
multiply by x^2
a(n-1)x^(n)+b(n-2)x^(n-1)+....=d
And the result is the same as the simplification as when you set MC=ATC, which proves that MC intersects ATC at the minimum of ATC. One could extend this to AVC by saying that since
TC = Variable Cost + Fixed Cost
Fixed cost is a constant so VC can be modeled as a polynomial
ax^n+bx^(n-1)+....+cx+d-alpha
where alpha is the fixed cost. Afterwards, you would use the same logic as above for ATC in order to prove the same for AVC.
Now my question is, how would this be proved for functions that were not polynomial in nature? Is it possible for total cost function models to be exponential, logarithmic, or trigonometric in nature? If so, I do not see an easy way of proving the same thing for this model.
An alternative is if you can prove the same for functions in general, but I can't see how one would approach this.
If anyone can think of the answer to this or can find it online somehow (I've tried but can't), it would be appreciated
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