I can get to the result by trying different values of X and Y but that is of course time taking. I want to know if there is a better way to get to the result?
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Have you heard of Lagrange multipliers? – beep-boop Oct 15 '14 at 17:32
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1Or write $Y = 10-X$, and differentiate. – Daniel Fischer Oct 15 '14 at 17:35
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You can probably use the Arithmetic geometric mean equality. We have: $$x+y=10$$ $$\frac{x}{3}+\frac{x}{3}+\frac{x}{3}+\frac{y}{2}+\frac{y}{2}=10$$ Since x and y are both positive, $$AM\ge GM$$ $$\frac{10}{5} \ge \left(\frac{x^3y^2}{3^32^2}\right)^\frac{1}{5}$$ $$\frac{x^3y^2}{3^32^2} \le 2^5$$ $$x^3y^2 \le 2^73^3$$
najayaz
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If, $y = 10 - x$ then the expression can be written as function of one variable: $$f(x) = x^3(10-x)^2$$ Now all you have to do is to differentiate this function and find its critical points.
Andrei Rykhalski
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There is a limitation on X and Y. On differentiating I get X = 16 as one critical point. But X can not be 16. – user2409011 Oct 15 '14 at 17:39
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@AndreiRykhalski What user2409011 was referring to was probably that it is given that x and y are both positive numbers. – Sherlock Holmes Oct 15 '14 at 22:37
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$Y = 10 - X$,and $X^3Y^2 = X^3(10-X)^2 = X^3(X^2 - 20X + 100) = X^5 - 20X^4 + 100X^3 = f(X)$.
$f'(X) = 5X^4 - 80X^3 + 300X^2 = 0 \iff 5X^2(X^2 - 16X + 60) = 5X^2(X-6)(X-10) = 0 \iff X = 0, 6, 10$. Since $0 < X < 10$, $X = 6$, and $Y = 10 - 6 = 4$. From this the max value of $X^3Y^2$ is $6^34^2 = 3,456$.
DeepSea
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You have $y=x-10$, so you're looking for maximum value of
$$x^3y^2=x^3(10-x)^2=100x^3+x^5-20x^4$$
When $x \to \infty$ then $100x^3+x^5-20x^4 \to \infty$, so there isn't maximal value of $x^3y^3$.
agha
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We just have to find the stationary points of $$ f(x) = x^3(10-x)^2 $$ whose derivative is: $$ f'(x) = 5x^2(10-x)(6-x) $$ hence the maximum of $x^3 y^2$ when $x+y=10$ is attained at $x=6$.
Jack D'Aurizio
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