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Question Number 12796 by tawa last updated on 01/May/17
The sum of two positive  numbers is 20. find the numbers  (i)  If their product is maximum  (ii)  If the sum of their square is maximum  (iii) If the product of the square of one and the cube of the other is maximum
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{positive}\:\:\mathrm{numbers}\:\mathrm{is}\:\mathrm{20}.\:\mathrm{find}\:\mathrm{the}\:\mathrm{numbers} \\ $$$$\left(\mathrm{i}\right)\:\:\mathrm{If}\:\mathrm{their}\:\mathrm{product}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\left(\mathrm{ii}\right)\:\:\mathrm{If}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{their}\:\mathrm{square}\:\mathrm{is}\:\mathrm{maximum} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{If}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{one}\:\mathrm{and}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{of}\:\mathrm{the}\:\mathrm{other}\:\mathrm{is}\:\mathrm{maximum} \\ $$
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⇒a=b=10  or:  f(a)=a(20−a)=20a−a^2   df/da=20−2a=0⇒a=10=b  ii)f(a)=a^2 +(20−a)^2 =2a^2 −40a+400  df/da=4a−40=0⇒a=10=b.  iii)f(a)=a^2 +(20−a)^3 or a^3 +(20−a)^2   ⇒df/da=[2a−3(20−a)^2 ]or [3a^2 −2(20−a)]  1)2a−3(400−40a+a^2 )=0  3a^2 −122a+1200=0  a=((122±(√(122^2 −4×3×1200)))/6)=((122±22)/6)=22,16.6  a=16.6,b=3.4  2)3a^2 −2(20−a)=0⇒3a^2 +2a−40=0  a=((−2±(√(4+3×4×40)))/6)=((−2±22)/6)=−4,3.3  a=3.3,b=16.7
$$\left.{i}\right){a}+{b}=\mathrm{20},{ab}={max}\Rightarrow{a}={b}=\mathrm{10} \\ $$$${or}: \\ $$$${f}\left({a}\right)={a}\left(\mathrm{20}−{a}\right)=\mathrm{20}{a}−{a}^{\mathrm{2}} \\ $$$${df}/{da}=\mathrm{20}−\mathrm{2}{a}=\mathrm{0}\Rightarrow{a}=\mathrm{10}={b} \\ $$$$\left.{ii}\right){f}\left({a}\right)={a}^{\mathrm{2}} +\left(\mathrm{20}−{a}\right)^{\mathrm{2}} =\mathrm{2}{a}^{\mathrm{2}} −\mathrm{40}{a}+\mathrm{400} \\ $$$${df}/{da}=\mathrm{4}{a}−\mathrm{40}=\mathrm{0}\Rightarrow{a}=\mathrm{10}={b}. \\ $$$$\left.{iii}\right){f}\left({a}\right)={a}^{\mathrm{2}} +\left(\mathrm{20}−{a}\right)^{\mathrm{3}} \boldsymbol{{or}}\:{a}^{\mathrm{3}} +\left(\mathrm{20}−{a}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{df}/{da}=\left[\mathrm{2}{a}−\mathrm{3}\left(\mathrm{20}−{a}\right)^{\mathrm{2}} \right]\boldsymbol{{or}}\:\left[\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{20}−{a}\right)\right] \\ $$$$\left.\mathrm{1}\right)\mathrm{2}{a}−\mathrm{3}\left(\mathrm{400}−\mathrm{40}{a}+{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\mathrm{3}{a}^{\mathrm{2}} −\mathrm{122}{a}+\mathrm{1200}=\mathrm{0} \\ $$$${a}=\frac{\mathrm{122}\pm\sqrt{\mathrm{122}^{\mathrm{2}} −\mathrm{4}×\mathrm{3}×\mathrm{1200}}}{\mathrm{6}}=\frac{\mathrm{122}\pm\mathrm{22}}{\mathrm{6}}=\mathrm{22},\mathrm{16}.\mathrm{6} \\ $$$$\boldsymbol{{a}}=\mathrm{16}.\mathrm{6},\boldsymbol{{b}}=\mathrm{3}.\mathrm{4} \\ $$$$\left.\mathrm{2}\right)\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{20}−{a}\right)=\mathrm{0}\Rightarrow\mathrm{3}{a}^{\mathrm{2}} +\mathrm{2}{a}−\mathrm{40}=\mathrm{0} \\ $$$${a}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{3}×\mathrm{4}×\mathrm{40}}}{\mathrm{6}}=\frac{−\mathrm{2}\pm\mathrm{22}}{\mathrm{6}}=−\mathrm{4},\mathrm{3}.\mathrm{3} \\ $$$$\boldsymbol{{a}}=\mathrm{3}.\mathrm{3},\boldsymbol{{b}}=\mathrm{16}.\mathrm{7} \\ $$
Commented by tawa last updated on 01/May/17
wow, God bless you sir.
$$\mathrm{wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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