Given that 2u+3v+4x is equal to 12. Then find the maximum value of u^2×v^3×x^5.

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- Oct 8th 2012, 01:07 AMgeniusgarvilmaximum value
Given that 2u+3v+4x is equal to 12. Then find the maximum value of u^2×v^3×x^5.

- Oct 8th 2012, 01:16 AMchiroRe: maximum value
Hey geniusgarvil.

If you perform a trinomial expansion on your 2u+4x+4x, you will be able to get a term in regards to u^2*v^3*x^5 if you raise it to the power of (2+3+5) = 10.

If you assume that the highest term contains all the information, then you can verify for yourself by setting one value to the highest and setting the rest equal to zero what the highest value of this term will be, by looking at the relevant coffecient that corresponds to that trinomial term.

So for a term of u^2*v^3*x^5, this will have a trinomial term in the same that a binomial has an nCr term before x^r*y(n-r) and this term is called the multinomial coeffecient:

Multinomial theorem - Wikipedia, the free encyclopedia - Oct 8th 2012, 01:47 AMMarkFLRe: maximum value
While I am aware that you posted in the Algebra forum, a method from differential calculus may be used, called Lagrange multipliers for optimization with a constraint.

If we define:

$\displaystyle f(u,v,x)=u^2v^3x^5$

subject to the constraint:

$\displaystyle g(u,v,x)=2u+3v+4x-12=0$

Then we find:

$\displaystyle 2uv^3x^5=2\lambda$

$\displaystyle 3u^2v^2x^5=3\lambda$

$\displaystyle 5u^2v^3x^4=4\lambda$

Using this system, along with the constraint, you can now determine the optimal values of the 3 variables. - Oct 8th 2012, 01:52 AMgeniusgarvilRe: maximum value
can you solve it in a simpler way. i m no pro in algebra

- Oct 8th 2012, 02:02 AMMarkFLRe: maximum value
What methods have you been using for similar problems? I would use the method of Lagrange, but I don't know what you are being taught to use.