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Question Number 110669 by Aina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Given}\mathrm{that}\mathrm{f}\left(\mathrm{x}\right)={(3+2\mathrm{x})}^3{(4-\mathrm{x})}^4\mathrm{on}$$ $$\mathrm{the}\mathrm{interval}-\frac{3}{2}<\mathrm{x}<4.\mathrm{Find}\mathrm{the}$$ $$\left(\mathrm{a}\right)\mathrm{Maximum}\mathrm{value}\mathrm{of}\mathrm{f}\left(\mathrm{x}\right)$$ $$\left(\mathrm{b}\right)\mathrm{The}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{that}\mathrm{gives}\mathrm{the}$$ $$\mathrm{maximum}\mathrm{in}\left(\mathrm{a}\right)$$

Commented byHer_Majesty last updated on 30/Aug/20

$${(3+2x)}^3(4-x^4)orrather{(3+2x)}^3{(4-x)}^4?$$

Commented byAina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{You}\mathrm{are}\mathrm{correct}.\mathrm{I}\mathrm{have}\mathrm{changed}\mathrm{it}.$$ $$\mathrm{Thanks}.$$

Answered by Her_Majesty last updated on 30/Aug/20

$$f^{\prime }\left(x\right)=6{(3+2x)}^2{(4-x)}^4-4{(4-x)}^3{(3+2x)}^3=$$ $$=2{(3+2x)}^2{(4-x)}^3(3(4-x)-2(3+2x))=$$ $$=2{(3+2x)}^2{(4-x)}^3(6-7x)$$ $$f^{\prime }\left(x\right)=0\Rightarrow x_{1,2}=-\frac{3}{2}\wedge x_{3,4,5}=4\wedge x_6=\frac{6}{7}$$ $$nowobviouslyf(-\frac{3}{2})=f\left(4\right)=0but$$ $$f\left(\frac{6}{7}\right)=\frac{2^4{3}^3{11}^7}{7^7}=\frac{8418457872}{823543}\approx 10222.2$$

Answered by 1549442205PVT last updated on 30/Aug/20

$$$$ $$\mathrm{Apply}{\mathrm{Cauchy}}^{\prime }\mathrm{s}\mathrm{inequality}\mathrm{we}\mathrm{have}$$ $${(3+2\mathrm{x})}^3{(4-\mathrm{x})}^4=\frac{{(3+2\mathrm{x})}^3{(6-1.5\mathrm{x})}^4}{1.5^4}\left(1\right)$$ $$\mathrm{Apply}{\mathrm{Cauchy}}^{\prime }\mathrm{s}\mathrm{inequality}\mathrm{we}\mathrm{have}$$ $$7{}^7\sqrt{(3+2\mathrm{x})(3+2\mathrm{x})(3+2\mathrm{x})(6-1.5\mathrm{x})(6-1.5\mathrm{x})(7-1.5\mathrm{x})(6-1.5\mathrm{x})}$$ $$⩽(3+2\mathrm{x})+(3+2\mathrm{x})+(3+2\mathrm{x})+(6-1.5\mathrm{x})+(6-1.5\mathrm{x})+(7-1.5\mathrm{x})+(6-1.5\mathrm{x})$$ $$=33\left(2\right)$$ $$\Rightarrow {(3+2\mathrm{x})}^3{(6-1.5\mathrm{x})}^4⩽{\left(\frac{33}{7}\right)}^7$$ $$\frac{{(3+2\mathrm{x})}^3{(6-1.5\mathrm{x})}^4}{1.5^4}⩽\frac{{33}^7}{7^7.1.5^4}$$ $$\mathrm{The}\mathrm{equality}\mathrm{ocurrs}\mathrm{if}\mathrm{and}\mathrm{only}\mathrm{if}$$ $$3+2\mathrm{x}=6-1.5\mathrm{x}\iff 3.5\mathrm{x}=3\iff \mathrm{x}=\frac{6}{7}$$ $$\mathrm{Thus}\mathrm{f}\left(\mathrm{x}\right)\mathrm{has}\mathrm{greatest}\mathrm{value}\mathrm{equal}\mathrm{to}$$ $$\frac{{33}^7}{7^7.1.5^4}\mathrm{when}\mathrm{x}=\frac{6}{7}$$

Commented byAina Samuel Temidayo last updated on 30/Aug/20

$$\mathrm{Yes}.\mathrm{Both}\mathrm{are}\mathrm{correct}.$$

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