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Question: How would one maximise $$ \small\frac{100!}{18!62!14!6!} (0.75x+0.2y)^{18} \left(-\frac{5x}{12}-\frac{y}{60}+\frac23\right)^{62} \left(\frac29-\frac{2x}{9}-\frac{341y}{3600}\right)^6 \left(\frac19-\frac{x}{9}-\frac{319y}{3600}\right)^{14} $$

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Consider using MathJax to make this more readable. –  robjohn Nov 19 '12 at 20:30
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That's an expression, not an equation. –  joriki Nov 19 '12 at 20:32
    
I added a right paren to make things match. –  robjohn Nov 19 '12 at 20:40

1 Answer 1

This expression has no maximum. If you put $$x=y=t$$

then the constant in front is positive, and each term in the product is of the form $(at+b)^k$ with $k$ even. So individually each term in the product approaches $+\infty$ with $t$ [for any particular $at+b$ the absolute value grows to infinity, and the even power on the $at+b$ keeps that behavior].

So unless the problem is rephrased with some kind of restriction on $x,y$ it looks like there is no maximum.

If there is some restriction made, one would start by looking at critical points, which given the form of the expression all have to lie on one or more of the lines given by setting $ax+by+c=0$. [EDIT: for each partial there is also a cubic factor coming from the products in threes of the linear factors muliplied by a coefficient.] This could reduce things to a number of one variable searches for the maximum on whatever the restricted region is. But the degree being high, this would be difficult to get exact maximum.

ADDED NOTE: The expression does have a minimum since it's bounded below by $0$; a search for that would again involve the critical points, and the above remarks would boil things down to several one variable min problems, but the high degree of the polynomials would likely make it unfeasible to look for the exact minimum; a numerical method might be best for getting a minimum.

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