Minima maxima: a²y = x⁴

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Francis June E....
Francis June E. Seraspe's picture
Minima maxima: a²y = x⁴

How to solve a²y = x⁴

Jhun Vert
Jhun Vert's picture

Your question don't have enough details. What are you trying to solve from this equation? Tagging it with Maxima and Minima is not enough information.

Francis June E....
Francis June E. Seraspe's picture

Yung Ans. Niya po is (0,0) at Minimum.
Yung process sana po sir. Thanks po

Francis June E....
Francis June E. Seraspe's picture

Yung process po sana kung paano solve.

Yung ans nya po is (0,0) , minimum.

Jhun Vert
Jhun Vert's picture

$a^2 y = x^4$
 

Differentiate
$a^2 y' = 4x^3$
 

Equate y' = 0 to determine the critical points (maxima or minima)
$a^2 (0) = 4x^3$

$x = 0$
 

For x = 0
$a^2 y = 0^4$

$y = 0$
 

Hence,
critical point = (0, 0)
 

Check the neighboring points to determine whether (0, 0) is minimum or maximum. Set x = ±1:
$a^2 y = (\pm 1)^4$

$y = +\dfrac{1}{a^2}$   ←   above (0, 0)
 

Hence, the point (0, 0) is minimum.

Francis June E....
Francis June E. Seraspe's picture

Salamat po sir

How about 9a²y=x(4a+x)³

Yung ans nya po is (a,3a) maximum

Thanks in advance sir

Jhun Vert
Jhun Vert's picture

Please create another forum post for your 2nd question as moderators won't allow multiple questions in one thread.

 
 
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