Question

If $a,b,c\in {{R}^{+}}$ such that $a+b+c=18$ , then find the maximum value of ${{a}^{2}}{{b}^{3}}{{c}^{4}}$ .

Answer 1

It is given that $A.M\ge G.M$ such that $a+b+c=18$ . We have to find the maximum value of the expression, ${{a}^{2}}{{b}^{3}}{{c}^{4}}$ . The exponents of $a,b,$ and $c$ are 2, 3, and 4 respectively. Now, divide $a,b,$ and $c$ by 2, 3, and 4 equal parts. We know the property that the arithmetic mean of all real positive numbers is greater than or equal to the geometric mean, $A.M\ge G.M$ . Now, apply the relation $A.M\ge G.M$ for the numbers $\dfrac{a}{2},\dfrac{a}{2},\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4}$ and solve it further to get the maximum value of the expression ${{a}^{2}}{{b}^{3}}{{c}^{4}}$ .

Complete step-by-step answer :
According to the question, it is given that we have three numbers $a,b,c$ such that $a,b,c\in {{R}^{+}}$ .
The summation of these three numbers is 18. So,
$a+b+c=18$ ……………………………………….(1)
We have to find the maximum value of ${{a}^{2}}{{b}^{3}}{{c}^{4}}$ ………………………………………………(2)
From equation (2). We have
The exponent of $a$ = 2 …………………………………………(3)
The exponent of $b$ = 3 …………………………………………(4)
The exponent of $c$ = 4 ………………………………………….(5)
Now, on dividing the number $a$ into equal parts as of its exponents, we get
$\dfrac{a}{2},\dfrac{a}{2}$ ………………………………………(6)
Similarly, on dividing the number $b$ into equal parts as of its exponents, we get
$\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3}$ ………………………………………(7)
Similarly, on dividing the number $c$ into equal parts as of its exponents, we get
$\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4}$ ………………………………………(8)
From equation (6), equation (7), and equation (8), we have the numbers
$\dfrac{a}{2},\dfrac{a}{2},\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4}$ …………………………………………….(9)
We know the property that the arithmetic mean of all real positive numbers is greater than or equal to the geometric mean, $A.M\ge G.M$ ……………………………………….(10)
It is given that $a,b,c$ are three numbers such that $a,b,c\in {{R}^{+}}$ .
So, the numbers $\dfrac{a}{2},\dfrac{a}{2},\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4}$ are also positive real numbers. Therefore, here we can apply the property shown in equation (10).
Now, on applying the relation $A.M\ge G.M$ for the positive real numbers $\dfrac{a}{2},\dfrac{a}{2},\dfrac{b}{3},\dfrac{b}{3},\dfrac{b}{3},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4},\dfrac{c}{4}$ , we get
$\dfrac{\dfrac{a}{2}+\dfrac{a}{2}+\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{b}{3}+\dfrac{c}{4}+\dfrac{c}{4}+\dfrac{c}{4}+\dfrac{c}{4}}{9}\ge {{\left( \dfrac{a}{2}\times \dfrac{a}{2}\times \dfrac{b}{3}\times \dfrac{b}{3}\times \dfrac{b}{3}\times \dfrac{c}{4}\times \dfrac{c}{4}\times \dfrac{c}{4}\times \dfrac{c}{4} \right)}^{\dfrac{1}{9}}}$
$\dfrac{a+b+c}{9}\ge {{\left( \dfrac{{{a}^{2}}}{4}\times \dfrac{{{b}^{3}}}{27}\times \dfrac{{{c}^{4}}}{{{4}^{4}}} \right)}^{\dfrac{1}{9}}}$ ………………………………………..(11)
From equation (1), we have the value of the expression $\left( a+b+c \right)$ .
Now, on substituting the expression $\left( a+b+c \right)$ by 18 in equation (11), we get

& \dfrac{18}{9}\ge {{\left( \dfrac{{{a}^{2}}}{4}\times \dfrac{{{b}^{3}}}{{{3}^{3}}}\times \dfrac{{{c}^{4}}}{{{4}^{4}}} \right)}^{\dfrac{1}{9}}} \\\ & 2\ge {{\left( \dfrac{{{a}^{2}}{{b}^{3}}{{c}^{4}}}{{{3}^{3}}\times {{4}^{5}}} \right)}^{\dfrac{1}{9}}} \\\ & {{2}^{9}}\ge \left( \dfrac{{{a}^{2}}{{b}^{3}}{{c}^{4}}}{{{3}^{3}}\times {{4}^{5}}} \right) \\\ & {{2}^{9}}\times {{3}^{3}}\times {{4}^{5}}\ge \left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right) \\\ & {{2}^{9}}\times {{3}^{3}}\times {{\left( {{2}^{2}} \right)}^{5}}\ge \left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right) \\\ & {{2}^{9}}\times {{3}^{3}}\times {{2}^{10}}\ge \left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right) \\\ & {{2}^{19}}{{3}^{3}}\ge \left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right) \\\ \end{aligned}$$ We can see that the expression $$\left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right)$$ is always less than or equal to $${{2}^{19}}{{3}^{3}}$$ . Therefore, when the expression $$\left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right)$$ has its value equal to $${{2}^{19}}{{3}^{3}}$$ , then it will have its maximum value. **Hence, the maximum value of the expression $$\left( {{a}^{2}}{{b}^{3}}{{c}^{4}} \right)$$ is $${{2}^{19}}{{3}^{3}}$$ .** **Note** : Whenever this type of question appears where we have to find the maximum value of an expression having real positive numbers. The best way to solve this question, first of all, divide the numbers into equal parts as of its exponents. Now, we know that $$A.M\ge G.M$$ for all real positive numbers. So, always think to apply the relation $$A.M\ge G.M$$ .