Minima maxima: a²y = x⁴

Francis June E. Seraspe's picture

How to solve a²y = x⁴

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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.

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Francis June E. Seraspe's picture

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

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Francis June E. Seraspe's picture

Yung process po sana kung paano solve.

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

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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.

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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

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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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