Question

Find A and B given that the function y=Ax−−√+Bx−−√ has a minimum value of 8 at x=1.

Answer 1

Answer:

$A=\pm 2\\B=\pm 2$

Step-by-step explanation:

Given that

$[tex]8=\sqrt{A}+\sqrt{A}\\\\8=2\sqrt{A}\\\\\sqrt{A}=4\\\\A=\pm 2\\\\which\,\, implies\\ \\B=\pm 2$---(1)\\[/tex]

To find minimum differentiate both sides w.r.to x

$\frac{dy}{dx}=\frac{d}{dx}\sqrt{Ax}+\frac{d}{dx}\sqrt{Bx}\\\\\frac{dy}{dx}=\frac{A}{\sqrt{Ax}}+\frac{B}{\sqrt{Bx}}\\\\At\,\, minima\,\,\frac{dy}{dx}=0\\\\\frac{A}{\sqrt{Ax}}+\frac{B}{\sqrt{Bx}}=0\\\\\frac{A\sqrt{Bx}+B\sqrt{Ax}}{x\sqrt{AB}}=0\\\\\frac{\sqrt{x}(A\sqrt{B}+B\sqrt{A})}{x\sqrt{AB}}=0\\\\\frac{(A\sqrt{B}+B\sqrt{A})}{\sqrt{ABx}}=0\\\\(A\sqrt{B}+B\sqrt{A})=0\\\\A\sqrt{B}=-B\sqrt{A}\\\\\sqrt{A}=-\sqrt{B}\\\\A=B--(2)$

Substituting (2) in (1) at x=1, y=8

$8=\sqrt{A}+\sqrt{A}\\\\2\sqrt{A}=8\\\\\sqrt{A}=4\\\\A=\pm 2\\\\which\,\, implies\\\\B=\pm 2$