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Question Number 213417 by hardmath last updated on 04/Nov/24

$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\in \mathbb{N}$$$$\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}=63$$$$\mathrm{Find}:\mathrm{maksimum}(\mathrm{ab}+\mathrm{bc}+\mathrm{cd})=?$$

Answered by Frix last updated on 05/Nov/24

$$d=63-a-b-c$$$$f=ab-ac-c^2+63c$$$$\frac{df}{da}=b-c=0$$$$\frac{df}{db}=a=0$$$$\frac{df}{dc}=-a-2c+63=0$$$$a=d=0\wedge b=c=\frac{63}{2}$$$$ab+bc+cd={\left(\frac{63}{2}\right)}^2=\frac{3969}{4}$$

Commented by A5T last updated on 05/Nov/24

$$a,b,c,d\in \mathbb{N}\Rightarrow ab+bc+cd\in \mathbb{N}$$

Commented by Frix last updated on 05/Nov/24

$$\mathrm{Sorry}$$

Commented by Tinku Tara last updated on 05/Nov/24

$$\mathrm{You}\mathrm{answer}\mathrm{is}992,\mathrm{while}\mathrm{Mr}\mathrm{W}\mathrm{is}$$$$991.\mathrm{You}\mathrm{are}\mathrm{including}0\in \mathbb{N}?$$

Answered by mr W last updated on 05/Nov/24

$$ab+bc+cd=(a+c)(b+d)-ad$$$${\left(ad\right)}_{min}=1\times 1=1$$$$say(a+c)(b+d)=pq,withp+q=63$$$${\left[(a+c)(b+d)\right]}_{max}={\left(pq\right)}_{max}=31\times 32$$$$\Rightarrow {(ab+bc+cd)}_{max}=31\times 32-1=991$$

Commented by hardmath last updated on 05/Nov/24

$$\mathrm{thankyou}\mathrm{dearprofessor}$$

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