Answer
349.2k+ 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.
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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)