What is the sum of the intercepts on the axes | vedclass

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(D) Given the curve $\sqrt{x} + \sqrt{y} = 4$. Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} =

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$16$

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(D) Given the curve $\sqrt{x} + \sqrt{y} = 4$. Differentiating with respect to $x$,we get: $\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0$. $\frac{dy}{dx} = -\sqrt{\frac{y}{x}}$. At the point $(4, 4)$,the slope $m = -\sqrt{\frac{4}{4}} = -1$. The equation of the tangent at $(4, 4)$ is $y - 4 = -1(x - 4)$,which simplifies to $y - 4 = -x + 4$,or $x + y = 8$. To find the intercepts,we write the equation in intercept form: $\frac{x}{8} + \frac{y}{8} = 1$. The $x$-intercept is $8$ and the $y$-intercept is $8$. The sum of the intercepts is $8 + 8 = 16$.

What is the sum of the intercepts on the axes of the tangent to the curve $\sqrt{x} + \sqrt{y} = 4$ at the point $(4, 4)$?

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