If ab = 2a + 3b where a 0 and b 0,then th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $ab = 2a + 3b$. Since $a > 0$ and $b > 0$,we can express $b$ in terms of $a$:$b(a - 3) = 2a \implies b = \frac{2a}{a - 3}$.Since $b > 0$ and

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(B) Given $ab = 2a + 3b$. Since $a > 0$ and $b > 0$,we can express $b$ in terms of $a$:$b(a - 3) = 2a \implies b = \frac{2a}{a - 3}$.Since $b > 0$ and $a > 0$,we must have $a - 3 > 0$,so $a > 3$.Let $z = ab = a \left( \frac{2a}{a - 3} ight) = \frac{2a^2}{a - 3}$.To find the minimum,we differentiate $z$ with respect to $a$:$\frac{dz}{da} = \frac{(a - 3)(4a) - 2a^2(1)}{(a - 3)^2} = \frac{4a^2 - 12a - 2a^2}{(a - 3)^2} = \frac{2a^2 - 12a}{(a - 3)^2}$.Setting $\frac{dz}{da} = 0$ gives $2a(a - 6) = 0$. Since $a > 3$,we have $a = 6$.At $a = 6$,$b = \frac{2(6)}{6 - 3} = \frac{12}{3} = 4$.The value of $ab = 6 	imes 4 = 24$.Using the second derivative test or checking values around $a=6$,we confirm this is the minimum value.Thus,the minimum value of $ab$ is $24$.

If $ab = 2a + 3b$ where $a > 0$ and $b > 0$,then the minimum value of $ab$ is:

Similar Questions

If the sum of two numbers is $3$,find the maximum value of the product of the first number and the square of the second number.

The maximum value of the function $f(x)=2x^3-15x^2+36x-48$ on the set $A=\{x \mid x^2+20 \leq 9x\}$ is

The sum of two positive numbers is given. If the sum of their cubes is minimum,then

Consider a square of side $2 \ cm$. It is cut from one of its corners as shown in the adjacent figure. The maximum value of the sum of the perimeters of the two plane figures thus formed is

The shortest distance between the point $\left( \frac{3}{2}, 0 \right)$ and the curve $y = \sqrt{x}, (x > 0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module