If $a,b,c,d>0$ and $a+b+c+d=10$ then find the maximum value of $a{{b}^{2}}cd$ .
Answer
623.4k+ views
Hint: To solve this question we use the inequality of arithmetic mean and the geometric mean. It states that the arithmetic mean of given non-negative numbers is always greater than or equal to the geometric mean of the same numbers. The major condition is that they are equal if and only if the list is the same for both. We use this and then evaluate to find the maximum value.
Complete step by step solution:
Firstly, to find the maximum values we use the inequality of arithmetic mean and the geometric mean.
$AM\ge GM$
Upon representing this in a more understandable form,
We get, $\dfrac{{{a}_{1}}+{{a}_{2}}+.....{{a}_{n}}}{n}\ge \sqrt[n]{{{a}_{1}}{{a}_{2}}.....{{a}_{n}}}$
Now we have a, b, c, d as the variables. They are non-negative numbers.
Upon putting them in the inequality of arithmetic mean and the geometric mean we get,
$\Rightarrow \dfrac{a+\dfrac{b}{2}+\dfrac{b}{2}+c+d}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Now let us start evaluating the LHS first.
$\Rightarrow \dfrac{a+b+c+d}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
We are given from the question that, $a+b+c+d=10$
Let us now substitute the same in the LHS of our expression.
$\Rightarrow \dfrac{10}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Simplify the same.
$\Rightarrow 2\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Now let us evaluate the RHS side.
$\Rightarrow 2\ge \left( \dfrac{a{{b}^{2}}cd}{4} \right)$
Now let us multiply with four on both sides of the expression.
$\Rightarrow 2\times 4\ge \left( a{{b}^{2}}cd \right)$
$\Rightarrow 8\ge \left( a{{b}^{2}}cd \right)$
As we can see, $a{{b}^{2}}cd$ is less than or equal to $8$.
Hence, the maximum value of the same will be $8$.
Note: One must keep in mind that this inequality of arithmetic mean and the geometric mean should always be used on the same list. That is, the arithmetic mean of a list is greater than or equal to the geometric mean of the same list. Always try to solve in such a way that the given condition in the question may appear in any solution step and thereby substitute the values for easy evaluation.
Complete step by step solution:
Firstly, to find the maximum values we use the inequality of arithmetic mean and the geometric mean.
$AM\ge GM$
Upon representing this in a more understandable form,
We get, $\dfrac{{{a}_{1}}+{{a}_{2}}+.....{{a}_{n}}}{n}\ge \sqrt[n]{{{a}_{1}}{{a}_{2}}.....{{a}_{n}}}$
Now we have a, b, c, d as the variables. They are non-negative numbers.
Upon putting them in the inequality of arithmetic mean and the geometric mean we get,
$\Rightarrow \dfrac{a+\dfrac{b}{2}+\dfrac{b}{2}+c+d}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Now let us start evaluating the LHS first.
$\Rightarrow \dfrac{a+b+c+d}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
We are given from the question that, $a+b+c+d=10$
Let us now substitute the same in the LHS of our expression.
$\Rightarrow \dfrac{10}{5}\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Simplify the same.
$\Rightarrow 2\ge \left( a\times \dfrac{b}{2}\times \dfrac{b}{2}\times c\times d \right)$
Now let us evaluate the RHS side.
$\Rightarrow 2\ge \left( \dfrac{a{{b}^{2}}cd}{4} \right)$
Now let us multiply with four on both sides of the expression.
$\Rightarrow 2\times 4\ge \left( a{{b}^{2}}cd \right)$
$\Rightarrow 8\ge \left( a{{b}^{2}}cd \right)$
As we can see, $a{{b}^{2}}cd$ is less than or equal to $8$.
Hence, the maximum value of the same will be $8$.
Note: One must keep in mind that this inequality of arithmetic mean and the geometric mean should always be used on the same list. That is, the arithmetic mean of a list is greater than or equal to the geometric mean of the same list. Always try to solve in such a way that the given condition in the question may appear in any solution step and thereby substitute the values for easy evaluation.
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