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Question

Question: The minimum value of \[2x+3y\], when \[xy=6\] is (a) 9 (b) 12 (c) 8 (d) 6...

The minimum value of 2x+3y2x+3y, when xy=6xy=6 is
(a) 9
(b) 12
(c) 8
(d) 6

Explanation

Solution

For solving this question you should know about the general mathematical calculation of expressions. In this problem it is asked to find the values of x and y. And this can be solved by general addition and submission of any numbers or terms on both sides. As we can say any term is added or subtracted as a form of zero.

Complete step by step answer:
According to our question it is asked to us to find the minimum value of 2x+3y2x+3y, if xy=6xy=6.
Now, as we know that if we want to solve any equation or any expression then we can add any term as a form of zero. It means that there will be addition and submission of the same term with the same signs if we are adding or subtracting that from both sides of an equation. And if we add in only one side of the equation then we add and subtract the same term with one negative and one positive sign.
Let f(x)=2x+3yf\left( x \right)=2x+3y
f(x)=2x+18xf\left( x \right)=2x+\dfrac{18}{x} (xy=6\because xy=6 given)
On differentiating we get,
f(x)=218x2f'\left( x \right)=2-\dfrac{18}{{{x}^{2}}}
Put f(x)=0f'\left( x \right)=0 for maximum and minimum.

& \Rightarrow 0=2-\dfrac{18}{{{x}^{2}}} \\\ & \Rightarrow x=\pm 3 \\\ \end{aligned}$$ And $$f''\left( x \right)=\dfrac{36}{{{x}^{3}}}$$ $$\Rightarrow f''\left( x \right)=\dfrac{36}{{{x}^{3}}}>0$$ $$\therefore $$ At $$x=3$$, if x is minimum. The minimum value is $$f\left( 3 \right)=2\left( 3 \right)+3\left( 2 \right)$$ $$f\left( 3 \right)=12$$ **So, the correct answer is “Option b”.** **Note:** While solving this type of questions you have to always use the relation which is given in the question. Because that will help to solve them. And in equations we have to always add or subtract the term which will make it easy to the question.