Respuesta :
Answer:
Option e.
Step-by-step explanation:
Consider the given equation is
[tex]y=\dfrac{A}{\sqrt{x}}+B\sqrt{x}[/tex]
The minimum value of function is 54 at x=81.
Substitute y=54 and x=81 in the given function.
[tex]54=\dfrac{A}{\sqrt{81}}+B\sqrt{81}[/tex]
[tex]54=\dfrac{A}{9}+9B[/tex]
Multiply both sides by 9.
[tex]486=A+81B[/tex] .... (1)
Differentiate the given function with respect to x.
[tex]y'=\dfrac{-0.5A}{x^{3/2}}+\dfrac{0.5B}{\sqrt{x}}[/tex]
[tex]y'=\dfrac{0.5}{\sqrt{x}}(-\dfrac{A}{x}+B)[/tex]
Equate y'=0 t find the critical fpoint.
[tex]\dfrac{0.5}{\sqrt{x}}(-\dfrac{A}{x}+B)=0[/tex]
Substitute x=81 in the above equation.
[tex]\dfrac{0.5}{\sqrt{81}}(-\dfrac{A}{81}+B)=0[/tex]
[tex]-\dfrac{A}{81}+B=0[/tex]
[tex]-\dfrac{A}{81}=-B[/tex]
[tex]A=81B[/tex] .... (2)
Substitute this value in equation (1).
[tex]486=A+A[/tex]
[tex]486=2A[/tex]
[tex]243=A[/tex]
Substitute this value in equation (2).
[tex]243=81B[/tex]
[tex]3=B[/tex]
The value of A is 243 and the value of B is 3. Therefore, the correct option is e.
The value of A and B given that the function [tex]y=\frac{A}{\sqrt{x} }+B\sqrt{x}[/tex] has a minimum value of 54 at x = 81 is 243 and 3 respectively
Given the function
[tex]y=\frac{A}{\sqrt{x} }+B\sqrt{x}[/tex]
If y= 54 where x = 81, hence
[tex]54=\frac{A}{\sqrt{81} }+B\sqrt{81}\\54=\frac{A}{9}+9B\\486=A+81B\\ A+81B=486[/tex]
At the minimum point [tex]\frac{dy}{dx} = 0[/tex]
Differentiate the given function:
[tex]y=\frac{A}{\sqrt{x} }+B\sqrt{x}\\y'=\frac{-0.5A}{{x^{3/2}} }+\frac{B}{x^{1/2}} \\\frac{-0.5A}{{x^{3/2}} }+\frac{B}{x^{1/2}}=0[/tex]
Substitute x = 81 to hav:
[tex]\frac{-0.5A}{81^{2/3}} +\frac{B}{81^{1/2}}=0\\\frac{-A}{81} + B=0\\-A+81B=0\\A=81B ......................... 2[/tex]
Substitute equation 2 into 1:
[tex]81B+81B= 486\\162B=486\\B=\frac{486}{162} \\B=3[/tex]
Get the value of A:
[tex]A=81B\\A=81(3)\\A=243[/tex]
Hence the value of A and B given that the function [tex]y=\frac{A}{\sqrt{x} }+B\sqrt{x}[/tex] has a minimum value of 54 at x = 81 is 243 and 3 respectively
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