Question

If a,b,c∈r+ such that a + b + c = 18 then the maximum value of a2b3c4 is equal to

Answer 1

Keeping in view that a + b + c = 18
a 2b 3 c 4 is max.
The sum of factors is = a + b + c = 18
 hence product will be max. when all the factors are equal

a/2=b/3=c/4=a+b+c/2+3+4=18/9=2
a=4 b=6 c=8
hence max val 4^26^38^4=14155776

Answer 2

Concept

Arithmetic Progression (AP) may be a sequence of numbers so as, within which the difference between any two consecutive numbers may be a constant value. it's also called Arithmetic Sequence.

Given

we are providing $a,b,c\in R^{+}$ such that $a+b+c=18.$

Find

we have to search out the most value of $a^2 b^3 c^4$.

Solution

According to the question, it's providing we've got three numbers $a,b,c$

such that $a,b,c\in R^{+}$.

The summation of those three numbers is $18$. So,

$a+b+c=18$                           ……………………………………….(1)

We have to search out the most 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 amount  $a$into equal parts as of its exponents, we get

$\frac{a}{2}, \frac{a}{2}$………………………………………(6)

Similarly, on dividing the quotient $b$ into equal parts as of its exponents, we get

$\frac{b}{3}, \frac{b}{3},\frac{b}{3}$………………………………………(7)

Similarly, on dividing the quotient $c$ into equal parts as of its exponents, we get

$\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$………………………………………(8)

From equation (6), equation (7), and equation (8), we've the numbers$\frac{a}{2}, \frac{a}{2}, \frac{b}{3}, \frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$

…………………………………………….(9)

We know the property that the mean of all real positive numbers is larger than or adequate to the mean,$A.M\geq G.M$ ……………………………………….(10)

It is only if $a,b,c$are three numbers such $a,b,c\in R^+$.

So, the numbers $\frac{a}{2}, \frac{a}{2}, \frac{b}{3}, \frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$ also are positive real numbers. Therefore, here we will apply the property shown in equation (10).

Now, on applying the relation $A.M\geq G.M$ for the positive real numbers $\frac{a}{2}, \frac{a}{2}, \frac{b}{3}, \frac{b}{3},\frac{b}{3},\frac{c}{4},\frac{c}{4},\frac{c}{4},\frac{c}{4}$, we get

$\begin{aligned}\frac{\frac{a}{2}+\frac{a}{2}+\frac{b}{3}+\frac{b}{3}+\frac{b}{3}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4}}{9}&\geq \left(\frac{a}{2}\times\frac{a}{2}\times\frac{b}{3}\times\frac{b}{3}\times\frac{b}{3}\times\frac{c}{4}\times\frac{c}{4}\times\frac{c}{4}\times\frac{c}{4}\right)^{\frac{1}{9}}\\ \frac{a+b+c}{9}\geq \left(\frac{a^2}{4}\times \frac{b^3}{27}\times \frac{c^4}{4^4}\right)^{\frac{1}{9}}\end{aligned}$                      ………………………………………..(11)

From equation (1), we've the worth of the expression $(a+b+c)$.

Now, on substituting the expression $(a+b+c)$ by $18$ in equation (11), we get

$\begin{aligned}\frac{18}{9}& \geq\left(\frac{a^2}{4}\times\frac{b^3}{3^3}\times \frac{c^4}{4^4}\right)^{\frac{1}{9}}\\ 2&\geq\left(\frac{a^2 b^3 c^4}{3^4 \times 4^5}\right)^{\frac{1}{9}}\\ 2^{9}&\geq \left(\frac{a^2 b^3 c^4}{3^4 \times 4^5}\right)\\ 2^9\times 3^3\times 4^5&\geq (a^2b^3c^4)\\ 2^9\times 3^3\times (2^2)^5&\geq \left(a^2 b^3 c^4\right)\\ 2^{19}3^3&\geq(a^2b^3c^4)\end{aligned}$

We can see that the expression $a^2b^3c^4$ is often but or adequate to $21933$.

Therefore, when the expression $a^2b^3c^4$ has its value adequate to  $21933$, then it'll have its maximum value.

Hence, the utmost value of the expression $a^2 b^3 c^4$ is $21933$.

#SPJ2