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Find the greatest value of `x^2y^3, w h e r exa n dy`lie in the first quadrant on the line `3x+4y=5.` |
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Answer» Given that 3x + 4y = 5 since we have expression `x^(2) y^(3)` we consider `2((3x)/(2)) + 3 ((4y)/(3)) = 3x + 4y = 5` Using A.M `ge` G.M for weighted means, we get `(2((3x)/(2)) + 3 ((4)/(3)))/(2 + 3) ge [((3x)/(2))^(2) ((4y)/(3))^(3)]^((1)/(5))` `implies ((3x + 4y)/(5))^(5) ge ((3)/(2))^(2) ((4)/(3))^(3) x^(2) y^(3)` `implies x^(2) y^(2) le ((2)/(3))^(2) ((3)/(4))^(3)` `implies x^(2) y^(3) le (3)/(16)` so, the greatest value of `x^(2) y^(3)` is `(3)/(16)` |
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