# Let $2x+3y+4z=9$, x, y, z > $0$ then the maximum value of ${{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}}$ is

A) ${{2}^{8}}$

B) ${{4}^{9}}$

C) ${{\left( \dfrac{11}{3} \right)}^{9}}$

D) ${{\left( \dfrac{11}{6} \right)}^{9}}$

Answer

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**Hint:**The maximum value of any function is denoted at the values of variables x, y and z which makes the value of the function the highest value among all possible values of x, y and z.

The values of function to the left of maximum are rising while at the right, its values are falling.

**Complete step by step solution:**

It is given that a = $2x+3y+4z=9$. The maximum value of function, b = ${{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}}$ is to be determined.

As functions a and b are related in terms of powers and bases, it can be said that

$\Delta b=\lambda \Delta a$

$\lambda $ is an arbitrary value.

Δb is determined as:

$

b={{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}} \\

\Delta b=\dfrac{db}{dx},\dfrac{db}{dy},\dfrac{db}{dz} \\

\Delta b = 2(1+x){{(2+y)}^{3}}{{(4+z)}^{4}},3{{(1+x)}^{2}}{{(2+y)}^{2}}{{(4+z)}^{4}},4{{(1+x)}^{3}}{{(2+y)}^{3}}{{(4+z)}^{3}}

$

Δa is determined as:

$

a=2x+3y+4z \\

\Delta a=\dfrac{da}{dx},\dfrac{da}{dy},\dfrac{da}{dz} \\

\Delta a=2,3,4

$

By putting values of Δb and Δa in above relation,

$\begin{align}

& \Delta b=\lambda \Delta a \\

& \left[ 2(1+x){{(2+y)}^{3}}{{(4+z)}^{4}},3{{(1+x)}^{2}}{{(2+y)}^{2}}{{(4+z)}^{4}},4{{(1+x)}^{3}}{{(2+y)}^{3}}{{(4+z)}^{3}} \right]=\lambda [2,3,4] \\

\end{align}$

On comparing,

$\begin{align}

& \left[ 2(1+x){{(2+y)}^{3}}{{(4+z)}^{4}} \right]=\lambda \times 2 \\

& (1+x){{(2+y)}^{3}}{{(4+z)}^{4}}=\lambda \text{ -eq(1)}

\end{align}$

$\begin{align}

& 3{{(1+x)}^{2}}{{(2+y)}^{2}}{{(4+z)}^{4}}=\lambda \times 3 \\

& {{(1+x)}^{2}}{{(2+y)}^{2}}{{(4+z)}^{4}}=\lambda \text{ -eq(2)}

\end{align}$

$\begin{align}

& 4{{(1+x)}^{3}}{{(2+y)}^{3}}{{(4+z)}^{3}}=\lambda \times 4 \\

& {{(1+x)}^{3}}{{(2+y)}^{3}}{{(4+z)}^{3}}=\lambda \text{ -eq(3)}

\end{align}$

Solving three equations gives,

x = \[\dfrac{8}{3}\] , y = \[\dfrac{5}{3}\], z = \[\dfrac{-1}{3}\].

Maximum value of function b is calculated by putting the determined values of x, y and z at which function b is maximized.

Maximized value of function b = ${{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}}$ at x = \[\dfrac{8}{3}\] , y = \[\dfrac{5}{3}\], z = \[\dfrac{-1}{3}\]

Maximized value = ${{\left( 1+\dfrac{8}{3} \right)}^{2}}{{\left( 2+\dfrac{5}{3} \right)}^{3}}{{\left( 4+\dfrac{\left( -1 \right)}{3} \right)}^{4}}$ = ${{\left( \dfrac{11}{3} \right)}^{2}}{{\left( \dfrac{11}{3} \right)}^{3}}{{\left( \dfrac{11}{3} \right)}^{4}}={{\left( \dfrac{11}{3} \right)}^{2+3+4}}={{\left( \dfrac{11}{3} \right)}^{9}}$

This indicates that option $(3)$ is correct.

**Note:**

Maximum value of function is calculated by taking a derivative so as to find the number below which the value of function is rising and after which the value of function is falling.

Maximized function has a peak as is the peak of a mountain. Firstly, it rises, reaches the maximum and then falls.

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