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Question: Let \(2x+3y+4z=9\), x, y, z > \(0\) then the maximum value of \({{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{...

Let 2x+3y+4z=92x+3y+4z=9, x, y, z > 00 then the maximum value of (1+x)2(2+y)3(4+z)4{{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}} is
A) 28{{2}^{8}}
B) 49{{4}^{9}}
C) (113)9{{\left( \dfrac{11}{3} \right)}^{9}}
D) (116)9{{\left( \dfrac{11}{6} \right)}^{9}}

Explanation

Solution

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=92x+3y+4z=9. The maximum value of function, b = (1+x)2(2+y)3(4+z)4{{(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
Δb=λΔa\Delta b=\lambda \Delta a
λ\lambda is an arbitrary value.
Δb is determined as:
b=(1+x)2(2+y)3(4+z)4 Δb=dbdx,dbdy,dbdz Δb=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 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 Δa=dadx,dady,dadz Δa=2,3,4 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,
Δb=λΔa [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]=λ[2,3,4] \begin{aligned} & \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{aligned}
On comparing,
[2(1+x)(2+y)3(4+z)4]=λ×2 (1+x)(2+y)3(4+z)4=λ -eq(1)\begin{aligned} & \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{aligned}
3(1+x)2(2+y)2(4+z)4=λ×3 (1+x)2(2+y)2(4+z)4=λ -eq(2)\begin{aligned} & 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{aligned}
4(1+x)3(2+y)3(4+z)3=λ×4 (1+x)3(2+y)3(4+z)3=λ -eq(3)\begin{aligned} & 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{aligned}
Solving three equations gives,
x = 83\dfrac{8}{3} , y = 53\dfrac{5}{3}, z = −13\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{{(1+x)}^{2}}{{(2+y)}^{3}}{{(4+z)}^{4}} at x = 83\dfrac{8}{3} , y = 53\dfrac{5}{3}, z = −13\dfrac{-1}{3}
Maximized value = (1+83)2(2+53)3(4+(−1)3)4{{\left( 1+\dfrac{8}{3} \right)}^{2}}{{\left( 2+\dfrac{5}{3} \right)}^{3}}{{\left( 4+\dfrac{\left( -1 \right)}{3} \right)}^{4}} = (113)2(113)3(113)4=(113)2+3+4=(113)9{{\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)(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.