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Question Number 12796 by tawa last updated on 01/May/17

$$\mathrm{The}\mathrm{sum}\mathrm{of}\mathrm{two}\mathrm{positive}\mathrm{numbers}\mathrm{is}20.\mathrm{find}\mathrm{the}\mathrm{numbers}$$$$\left(\mathrm{i}\right)\mathrm{If}\mathrm{their}\mathrm{product}\mathrm{is}\mathrm{maximum}$$$$\left(\mathrm{ii}\right)\mathrm{If}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{their}\mathrm{square}\mathrm{is}\mathrm{maximum}$$$$\left(\mathrm{iii}\right)\mathrm{If}\mathrm{the}\mathrm{product}\mathrm{of}\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{one}\mathrm{and}\mathrm{the}\mathrm{cube}\mathrm{of}\mathrm{the}\mathrm{other}\mathrm{is}\mathrm{maximum}$$

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 01/May/17

$$i)a+b=20,ab=max\Rightarrow a=b=10$$$$or:$$$$f\left(a\right)=a(20-a)=20a-a^2$$$$df/da=20-2a=0\Rightarrow a=10=b$$$$ii)f\left(a\right)=a^2+{(20-a)}^2=2a^2-40a+400$$$$df/da=4a-40=0\Rightarrow a=10=b.$$$$iii)f\left(a\right)=a^2+{(20-a)}^3\mathit{o}\mathit{r}{a}^3+{(20-a)}^2$$$$\Rightarrow df/da=[2a-3{(20-a)}^2]\mathit{o}\mathit{r}[3a^2-2(20-a)]$$$$1)2a-3(400-40a+a^2)=0$$$$3a^2-122a+1200=0$$$$a=\frac{122\pm \sqrt{{122}^2-4\times 3\times 1200}}{6}=\frac{122\pm 22}{6}=22,16.6$$$$\mathit{a}=16.6,\mathit{b}=3.4$$$$2)3a^2-2(20-a)=0\Rightarrow 3a^2+2a-40=0$$$$a=\frac{-2\pm \sqrt{4+3\times 4\times 40}}{6}=\frac{-2\pm 22}{6}=-4,3.3$$$$\mathit{a}=3.3,\mathit{b}=16.7$$

Commented by tawa last updated on 01/May/17

$$\mathrm{wow},\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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