Question

Find A and B given that the function y=Ax√+Bx√ has a minimum value of 54 at x = 81.

Answer 1

Answer:

Option e.

Step-by-step explanation:

Consider the given equation is  

$y=\dfrac{A}{\sqrt{x}}+B\sqrt{x}$

The minimum value of function is 54 at x=81.

Substitute y=54 and x=81 in the given function.

$54=\dfrac{A}{\sqrt{81}}+B\sqrt{81}$

$54=\dfrac{A}{9}+9B$

Multiply both sides by 9.

$486=A+81B$              .... (1)

Differentiate the given function with respect to x.

$y'=\dfrac{-0.5A}{x^{3/2}}+\dfrac{0.5B}{\sqrt{x}}$

$y'=\dfrac{0.5}{\sqrt{x}}(-\dfrac{A}{x}+B)$

Equate y'=0 t find the critical fpoint.

$\dfrac{0.5}{\sqrt{x}}(-\dfrac{A}{x}+B)=0$

Substitute x=81 in the above equation.

$\dfrac{0.5}{\sqrt{81}}(-\dfrac{A}{81}+B)=0$

$-\dfrac{A}{81}+B=0$

$-\dfrac{A}{81}=-B$

$A=81B$            .... (2)

Substitute this value in equation (1).

$486=A+A$

$486=2A$

$243=A$

Substitute this value in equation (2).

$243=81B$

$3=B$

The value of A is 243 and the value of B is 3. Therefore, the correct option is e.