Courses
Courses for Kids
Free study material
Offline Centres
More
Last updated date: 07th Dec 2023
Total views: 279.9k
Views today: 4.79k

# Let $2x+3y+4z=9$, x, y, z > $0$ then the maximum value of ${{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}}$ isA) ${{2}^{8}}$ B) ${{4}^{9}}$ C) ${{\left( \dfrac{11}{3} \right)}^{9}}$ D) ${{\left( \dfrac{11}{6} \right)}^{9}}$

Verified
279.9k+ views
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.