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Question Number 139505 by bemath last updated on 28/Apr/21

Find the greatest value of x^2 y^3    when 3x+4y=5

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{3}} \: \\ $$$$\mathrm{when}\:\mathrm{3x}+\mathrm{4y}=\mathrm{5} \\ $$

Commented by mr W last updated on 28/Apr/21

3x+4y=5 ⇒y∈(−∞,∞)  x^2 y^3 →−∞ if y→−∞  x^2 y^3 →+∞ if y→+∞  therefore x^2 y^3  has no minimum   and no maximum!

$$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{5}\:\Rightarrow{y}\in\left(−\infty,\infty\right) \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} \rightarrow−\infty\:{if}\:{y}\rightarrow−\infty \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} \rightarrow+\infty\:{if}\:{y}\rightarrow+\infty \\ $$$${therefore}\:{x}^{\mathrm{2}} {y}^{\mathrm{3}} \:{has}\:{no}\:{minimum}\: \\ $$$${and}\:{no}\:{maximum}! \\ $$

Answered by MJS_new last updated on 28/Apr/21

y=((5−3x)/4)  x^2 y^3 =−((27)/4)x^5 +((135)/4)x^4 −((225)/4)x^3 +((125)/4)x^2   this has got  a local minimum at x=0  a local maximum at x=(2/3)  but  −∞<x^2 y^3 <+∞

$${y}=\frac{\mathrm{5}−\mathrm{3}{x}}{\mathrm{4}} \\ $$$${x}^{\mathrm{2}} {y}^{\mathrm{3}} =−\frac{\mathrm{27}}{\mathrm{4}}{x}^{\mathrm{5}} +\frac{\mathrm{135}}{\mathrm{4}}{x}^{\mathrm{4}} −\frac{\mathrm{225}}{\mathrm{4}}{x}^{\mathrm{3}} +\frac{\mathrm{125}}{\mathrm{4}}{x}^{\mathrm{2}} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{got} \\ $$$$\mathrm{a}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{at}\:{x}=\mathrm{0} \\ $$$$\mathrm{a}\:\mathrm{local}\:\mathrm{maximum}\:\mathrm{at}\:{x}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\mathrm{but} \\ $$$$−\infty<{x}^{\mathrm{2}} {y}^{\mathrm{3}} <+\infty \\ $$

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