Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 177579 by mr W last updated on 07/Oct/22

$$if{x}^3+y^3+\frac{x+y}{4}=\frac{15}{2},findmaximum$$$$valueofx+y.$$

Answered by TheHoneyCat last updated on 07/Oct/22

$$\mathrm{if}x=y+ϵ$$$$15/2=x^3+y^3+\frac{x+y}{4}$$$$={(y+ϵ)}^3+y^3+\frac{2y+ϵ}{4}$$$$=y^3+3y^2ϵ+3y{ϵ}^2+{ϵ}^3+y^3+\frac{2y+ϵ}{4}$$$$=y^3+y^3+\frac{y+y}{4}+ϵ(3y^2+1/4)+{ϵ}^2\left(3y\right)$$$$$$$$Thefunctioninybeeingmonotonous$$$$\mathrm{This}\mathrm{shows}\mathrm{that}ϵ\mathrm{needs}\mathrm{to}\mathrm{be}\mathrm{decreese}(\mathrm{when}$$$$y\mathrm{is}\mathrm{increesed})$$$$\mathrm{if}\mathrm{you}\mathrm{are}\mathrm{to}\mathrm{maintain}\mathrm{the}\mathrm{equation}.$$$$$$$$\mathrm{Basicaly}‘‘x+y^{″}\mathrm{is}\mathrm{always}\mathrm{greater}\mathrm{when}ϵ=0.$$$$$$$$\mathrm{So}\mathrm{the}\mathrm{problem}\mathrm{boils}\mathrm{down}\mathrm{to}\mathrm{finding}\mathrm{the}$$$$\max \mathrm{value}\mathrm{of}2x$$$$\mathrm{when}$$$$2x^3+\frac{x}{2}=\frac{15}{2}$$$$\iff 4x^3+x=15$$$$\iff (x-3/2)(4x^2+6x+10)=0$$$$\mathrm{Since}{6}^2-4\times 10<0$$$$\mathrm{The}\mathrm{only}\mathrm{possible}\mathrm{solution}\mathrm{is}\mathrm{thus}:$$$$x=3/2$$$$$$$$(\mathrm{one}\mathrm{can}\mathrm{verify}\mathrm{that}\mathrm{it}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{minimum}$$$$\mathrm{by}\mathrm{comparing}\mathrm{to}\mathrm{values}\mathrm{for}x=0\mathrm{and}\mathrm{y}\ne 0...)$$

Commented by mr W last updated on 07/Oct/22

$$thankssir!$$

Answered by mr W last updated on 07/Oct/22

$$lett=x+y>0,s=xy⩽\frac{{(x+y)}^2}{4}=\frac{t^2}{4}$$$${(x+y)}^3-3xy(x+y)+\frac{x+y}{4}=\frac{15}{2}$$$$t^3-3st+\frac{t}{4}=\frac{15}{2}$$$$t^3+\frac{t}{4}=\frac{15}{2}+3st⩽\frac{15}{2}+\frac{3t^3}{4}$$$$\frac{t^3}{4}+\frac{t}{4}⩽\frac{15}{2}$$$$t^3+t⩽30$$$$t(t^2+1)⩽30$$$$\Rightarrow t⩽3\Rightarrow {(x+y)}_{max}=3$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com