The maximum value of xy when x + 2y = 8 is | vedclass

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

Question

(D) Given the constraint $x + 2y = 8$,we can express $y$ in terms of $x$ as $y = \frac{8 - x}{2}$.Let the function to be maximized be $f(x) = xy$. Sub

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Given the constraint $x + 2y = 8$,we can express $y$ in terms of $x$ as $y = \frac{8 - x}{2}$.Let the function to be maximized be $f(x) = xy$. Substituting the expression for $y$,we get $f(x) = x \left( \frac{8 - x}{2} ight) = 4x - \frac{x^2}{2}$.To find the critical points,we calculate the first derivative: $f'(x) = \frac{d}{dx} (4x - \frac{x^2}{2}) = 4 - x$.Setting $f'(x) = 0$,we get $4 - x = 0$,which implies $x = 4$.Substituting $x = 4$ into the constraint equation,we find $y = \frac{8 - 4}{2} = 2$.To verify that this is a maximum,we check the second derivative: $f''(x) = -1$. Since $f''(x) The maximum value is $f(4) = 4 	imes 2 = 8$.

The maximum value of $xy$ when $x + 2y = 8$ is

Similar Questions

What is the primary purpose of data entry in the context of mathematical analysis?

An inclined plane makes an angle of $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to ........

The surface of the lake is at $2^{\circ}C$. Find the temperature of the bottom of the lake in $^{\circ}C$.

The number of sigma $(\sigma)$ bonds in $but-1-ene$ is:

If $\cos 2x = (\sqrt{2} + 1) \left(\cos x - \frac{1}{\sqrt{2}}\right)$ and $\cos x \neq \frac{1}{\sqrt{2}}$,then $x \in$

Vedclass Test Series

Exam Paper Generator

Online Exam Module