The maximum value of xy subject to x + y = 8 | vedclass

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

Question

(B) Given $x + y = 8$,we can write $y = 8 - x$. Let $f(x) = xy = x(8 - x) = 8x - x^2$. To find the maximum value,we differentiate $f(x)$ with respect

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) Given $x + y = 8$,we can write $y = 8 - x$. Let $f(x) = xy = x(8 - x) = 8x - x^2$. To find the maximum value,we differentiate $f(x)$ with respect to $x$: $f'(x) = 8 - 2x$. Setting $f'(x) = 0$ for critical points: $8 - 2x = 0 \implies x = 4$. Since $f''(x) = -2 Substituting $x = 4$ into the equation $y = 8 - x$,we get $y = 4$. Therefore,the maximum value of $xy = 4 	imes 4 = 16$.

The maximum value of $xy$ subject to $x + y = 8$ is

Similar Questions

On the interval $[0,1]$,the function $f(x) = x^{25}(1-x)^{75}$ takes its maximum value at the point

Find the local maximum and local minimum values for the function $f(x) = \sin x - \cos x$ where $0 < x < 2\pi$.

Let $f(x)=3 \sqrt{x-2}+\sqrt{4-x}$ be a real-valued function. If $\alpha$ and $\beta$ are respectively the minimum and the maximum values of $f$,then $\alpha^2+2 \beta^2$ is equal to

Find the minimum value of $2x + 3y$ when $xy = 6$.

The sum of the maximum and the minimum values of $f(x) = 3x^4 - 2x^3 - 6x^2 + 6x + 4$ in the interval $(0, 2)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module