Find the least value of $3x + 4y$if ${x^2}{y^3} = 6$.
Last updated date: 13th Mar 2023
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Answer
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Hint: In the above question it is given that ${x^2}{y^3}$. So on the basis of this we will break down the term of $x$ in twice as there is ${x^2}$. So similarly we will break the term of $y$ in three parts as it is given that ${y^3}$. Now we have to find the minimum value of $3x + 4y$. So we will use the property of arithmetic mean and geometric mean to solve this question.
Complete step-by-step answer:
We will break the term $3x$ into $\dfrac{{3x}}{2},\dfrac{{3x}}{2}$ and we will break the another term $4y = \dfrac{{4y}}{3},\dfrac{{4y}}{3},\dfrac{{4y}}{3}$.
Now we will use the property of arithmetic mean (AM) and geometric mean (GM) . It says that $AM \geqslant GM$.
We know that formula of arithmetic mean is i.e. $\overline X = \dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{N}$, where $N$ is the total number of observations, here we have $N = 5$. Now applying the formula we solve it.
So we have AM of the above terms is
$AM = \dfrac{{\dfrac{{3x}}{2} + \dfrac{{3x}}{2} + \dfrac{{4y}}{3} + \dfrac{{4y}}{3} + \dfrac{{4y}}{3}}}{5}$.
We have to add all these terms so, by taking the LCM of the denominator and adding we have the value
$AM = \dfrac{{\dfrac{{9x + 9x + 8y + 8y + 8y}}{6}}}{5} = \dfrac{{\dfrac{{6(3x + 4y)}}{6}}}{5}$.
We have taken the common factor out in the numerator and then it gets cancelled with the denominator, so it gives us $\dfrac{{3x + 4y}}{5}$.
Now we will calculate the value of
GM$ = {\left( {\dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{4}{3} \times \dfrac{4}{3} \times \dfrac{4}{3}} \right)^{\dfrac{1}{5}}}$. On further solving we have ${(2 \times 2 \times 2 \times 2 \times 2)^{\dfrac{1}{5}}} = {2^{5 \times \dfrac{1}{5}}}$.
It gives us the value $GM = 2$.
Now by putting the values back in the formula we have: $\dfrac{{3x + 4y}}{5} \geqslant 2$, on solving we have $3x + 4y \geqslant 2 \times 5 \Rightarrow 3x + 4y \geqslant 10$. Here we can see that the value of $3x + 4y$ has to be greater or equal to $10$, it cannot be less than it.
Hence we can say that the minimum value of $3x + 4y$ is $10$.
So, the correct answer is “10”.
Note: We should know that if in AM, there is set of numbers ${a_1},{a_2},{a_3}...{a_n}$, then the value of AM is $\left( {\dfrac{{{a_1} + {a_2} + {a_3} + ...{a_n}}}{n}} \right)$, where $n$ is the number of terms. Similarly in G.M if set of values as ${a_1},{a_2},{a_3}...{a_n}$, then the $GM = {\left( {{a_1} \times {a_2} \times {a_3}...{a_n}} \right)^{\dfrac{1}{n}}}$, where $n$ is the number of terms. There is an inequality relation between the AM and GM , greater than equal to i.e. $AM \geqslant GM$.
Complete step-by-step answer:
We will break the term $3x$ into $\dfrac{{3x}}{2},\dfrac{{3x}}{2}$ and we will break the another term $4y = \dfrac{{4y}}{3},\dfrac{{4y}}{3},\dfrac{{4y}}{3}$.
Now we will use the property of arithmetic mean (AM) and geometric mean (GM) . It says that $AM \geqslant GM$.
We know that formula of arithmetic mean is i.e. $\overline X = \dfrac{{\sum\limits_{i = 1}^n {{X_i}} }}{N}$, where $N$ is the total number of observations, here we have $N = 5$. Now applying the formula we solve it.
So we have AM of the above terms is
$AM = \dfrac{{\dfrac{{3x}}{2} + \dfrac{{3x}}{2} + \dfrac{{4y}}{3} + \dfrac{{4y}}{3} + \dfrac{{4y}}{3}}}{5}$.
We have to add all these terms so, by taking the LCM of the denominator and adding we have the value
$AM = \dfrac{{\dfrac{{9x + 9x + 8y + 8y + 8y}}{6}}}{5} = \dfrac{{\dfrac{{6(3x + 4y)}}{6}}}{5}$.
We have taken the common factor out in the numerator and then it gets cancelled with the denominator, so it gives us $\dfrac{{3x + 4y}}{5}$.
Now we will calculate the value of
GM$ = {\left( {\dfrac{3}{2} \times \dfrac{3}{2} \times \dfrac{4}{3} \times \dfrac{4}{3} \times \dfrac{4}{3}} \right)^{\dfrac{1}{5}}}$. On further solving we have ${(2 \times 2 \times 2 \times 2 \times 2)^{\dfrac{1}{5}}} = {2^{5 \times \dfrac{1}{5}}}$.
It gives us the value $GM = 2$.
Now by putting the values back in the formula we have: $\dfrac{{3x + 4y}}{5} \geqslant 2$, on solving we have $3x + 4y \geqslant 2 \times 5 \Rightarrow 3x + 4y \geqslant 10$. Here we can see that the value of $3x + 4y$ has to be greater or equal to $10$, it cannot be less than it.
Hence we can say that the minimum value of $3x + 4y$ is $10$.
So, the correct answer is “10”.
Note: We should know that if in AM, there is set of numbers ${a_1},{a_2},{a_3}...{a_n}$, then the value of AM is $\left( {\dfrac{{{a_1} + {a_2} + {a_3} + ...{a_n}}}{n}} \right)$, where $n$ is the number of terms. Similarly in G.M if set of values as ${a_1},{a_2},{a_3}...{a_n}$, then the $GM = {\left( {{a_1} \times {a_2} \times {a_3}...{a_n}} \right)^{\dfrac{1}{n}}}$, where $n$ is the number of terms. There is an inequality relation between the AM and GM , greater than equal to i.e. $AM \geqslant GM$.
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