ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: ally1784 on February 27, 2021, 11:11:19 pm
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Someone please help me?
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Usually \(\frac{x+y+z}{3} \geq (xyz)^{\frac{1}{3}}\) with equality only holding under the conditions stated. If that sum on the left hand side is to have a minimum value, it is implied that \(x, y, z\) are all equal and hence \(3(xyz)^{\frac{1}{3}}\) can be computed using the product of three of \(x, y, z\) (with repetition). However, they've conveniently given that \(\frac{24}{a} = \frac{6}{b} = 12ab\) - what happens when you take the product \(\frac{24}{a} \times \frac{6}{b} \times 12ab\)? Try this out and see if you can understand what's happening - any more questions, let us know! :)
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so afterwards, how do I find value of "a" & "b"?
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If you've worked out the value of \(x, y, z\) then the solutions for a and b should be relatively trivial if you realise that \(x = y = z\). ie. \(\frac{72}{a} = \frac{18}{b} = 36ab = 3(xyz)^{\frac{1}{3}}\), since that is the constraint that gives the original equation equality. Subbing in \(x = \frac{24}{a}\) or \(x = \frac{6}{b}\) should yield an equation of one variable.
Hope this helps :)
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this was the original question