Question Number 62176 by Tawa1 last updated on 16/Jun/19
Answered by MJS last updated on 17/Jun/19
$$\mathrm{to}\:\mathrm{make}\:\mathrm{it}\:\mathrm{clear}\:\mathrm{let}:\:{a},\:{b},\:{c},\:{d}\:\in\mathbb{R} \\ $$$${a}+{b}+{c}+{d}=\mathrm{1} \\ $$$${F}={ab}+{bc}+{cd} \\ $$$$ \\ $$$${d}=\mathrm{1}−{a}−{b}−{c} \\ $$$${F}={ab}+{c}−{ac}−{c}^{\mathrm{2}} \\ $$$$\frac{{dF}}{{da}}={b}−{c}=\mathrm{0}\:\Rightarrow\:{c}={b} \\ $$$$\frac{{dF}}{{db}}={a}=\mathrm{0}\:\Rightarrow\:{a}=\mathrm{0} \\ $$$$\frac{{dF}}{{dc}}=\mathrm{1}−{a}−\mathrm{2}{c}=\mathrm{0}\:\Rightarrow\:{c}=\frac{\mathrm{1}−{a}}{\mathrm{2}}\:\Rightarrow\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{d}=\mathrm{0} \\ $$$${F}={ab}+{bc}+{cd}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same}\:\mathrm{on}\:\mathrm{a}\:\mathrm{different}\:\mathrm{path} \\ $$$${c}=\mathrm{1}−{a}−{b}−{d} \\ $$$${F}=\left({b}+{d}\right)−\left({b}+{d}\right)^{\mathrm{2}} −{ad} \\ $$$$\frac{{dF}}{{da}}=−{d}=\mathrm{0}\:\Rightarrow\:{d}=\mathrm{0} \\ $$$$\frac{{dF}}{{db}}=\mathrm{1}−\mathrm{2}\left({b}+{d}\right)=\mathrm{0}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}}−{d}\:\Rightarrow\:{b}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{dF}}{{dd}}=\mathrm{1}−\mathrm{2}\left({b}+{d}\right)−{a}=\mathrm{0}\:\Rightarrow\:{a}=\mathrm{0} \\ $$$$\Rightarrow\:{c}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{this}\:\mathrm{means}:\:{b}={c}\:\mathrm{must}\:\mathrm{be}\:\mathrm{as}\:\mathrm{big}\:\mathrm{and}\:{a}={d}\:\mathrm{as} \\ $$$$\mathrm{small}\:\mathrm{as}\:\mathrm{possible} \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$${a},\:{b},\:{c},\:{d}\:\in\mathbb{N}^{\bigstar} \\ $$$${a}+{b}+{c}+{d}=\mathrm{63} \\ $$$${a}={d}=\mathrm{1} \\ $$$${b}=\mathrm{31}\wedge{c}=\mathrm{30}\:\vee\:{b}=\mathrm{30}\wedge{c}=\mathrm{31} \\ $$$${F}=\mathrm{1}×\mathrm{31}+\mathrm{31}×\mathrm{30}+\mathrm{30}×\mathrm{1}=\mathrm{991} \\ $$$$ \\ $$$$\mathrm{if}\:{a},\:{b},\:{c},\:{d}\in\mathbb{R} \\ $$$${a}={d}=\mathrm{0} \\ $$$${b}={c}=\mathrm{31}.\mathrm{5} \\ $$$${F}=\mathrm{31}.\mathrm{5}^{\mathrm{2}} =\mathrm{992}.\mathrm{25} \\ $$
Commented by Tawa1 last updated on 21/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 21/Jun/19
![a+b+c+d=63 S=ab+bc+cd =ab+bc+cd+da−da =(a+c)(b+d)−da =(a+c)[63−(a+c)]−da to get max. S, d must be as small as possible which is 1. ⇒a+c=62−b S=(62−b)(1+b)−a to get max. S, a must be as small as possible which is 1. ⇒b+c=61⇒1≤b≤60 S=(62−b)(1+b)−1=61+(((61)/2))^2 −(b−((61)/2))^2 to get max. S, b should be as close to ((61)/2) as possible, which is 30 or 31. with b=30: S=32×31−1=991 with b=31: S=31×32−1=991 ⇒max. S=991](https://www.tinkutara.com/question/Q62446.png)
$${a}+{b}+{c}+{d}=\mathrm{63} \\ $$$${S}={ab}+{bc}+{cd} \\ $$$$={ab}+{bc}+{cd}+{da}−{da} \\ $$$$=\left({a}+{c}\right)\left({b}+{d}\right)−{da} \\ $$$$=\left({a}+{c}\right)\left[\mathrm{63}−\left({a}+{c}\right)\right]−{da} \\ $$$${to}\:{get}\:{max}.\:{S},\:{d}\:{must}\:{be}\:{as}\:{small}\:{as} \\ $$$${possible}\:{which}\:{is}\:\mathrm{1}. \\ $$$$\Rightarrow{a}+{c}=\mathrm{62}−{b} \\ $$$${S}=\left(\mathrm{62}−{b}\right)\left(\mathrm{1}+{b}\right)−{a} \\ $$$${to}\:{get}\:{max}.\:{S},\:{a}\:{must}\:{be}\:{as}\:{small}\:{as} \\ $$$${possible}\:{which}\:{is}\:\mathrm{1}. \\ $$$$\Rightarrow{b}+{c}=\mathrm{61}\Rightarrow\mathrm{1}\leqslant{b}\leqslant\mathrm{60} \\ $$$${S}=\left(\mathrm{62}−{b}\right)\left(\mathrm{1}+{b}\right)−\mathrm{1}=\mathrm{61}+\left(\frac{\mathrm{61}}{\mathrm{2}}\right)^{\mathrm{2}} −\left({b}−\frac{\mathrm{61}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${to}\:{get}\:{max}.\:{S},\:{b}\:{should}\:{be}\:{as}\:{close} \\ $$$${to}\:\frac{\mathrm{61}}{\mathrm{2}}\:{as}\:{possible},\:{which}\:{is}\:\mathrm{30}\:{or}\:\mathrm{31}. \\ $$$${with}\:{b}=\mathrm{30}:\:{S}=\mathrm{32}×\mathrm{31}−\mathrm{1}=\mathrm{991} \\ $$$${with}\:{b}=\mathrm{31}:\:{S}=\mathrm{31}×\mathrm{32}−\mathrm{1}=\mathrm{991} \\ $$$$\Rightarrow{max}.\:{S}=\mathrm{991} \\ $$
Commented by Tawa1 last updated on 21/Jun/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$