# Minima Maxima: 9a³y=x(4a-x)³

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Francis June E.... Minima Maxima: 9a³y=x(4a-x)³

Patulong po ulit ako mam/sir
Yung process po sana ulit
9a²y=x(4a+x)³
Yung ans nya po is (a,3a) maximum
Thanks in advance mam/sir

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Jhun Vert I think something is wrong with your given. If you check for x = a, y ≠ 3a. Here is the checking of your answer key.

Set x = a and solve for y
$9a^2 y = a(4a + a)^3$

$9a^2 y = 125a^4$

$y = \dfrac{125a^2}{9}$   ←   not equal to 3a

Kindly check the given.

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Francis June E.... Ay sorry po.
Eto po sir 9a²y=x(4a-x)³

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Jhun Vert The same error. You can evaluate if x = a, y = 3a2 not 3a. Either the given is wrong or the answer key is wrong.

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Francis June E.... Kinuha ko kasi sa libro ni Love and Rainville yang example sir.

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Jhun Vert i-check ko. Anong page?

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Francis June E.... 62-64 sir

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Jhun Vert Okay I got it. The given is actually $9a^3 y = x(4a - x)^3$

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Jhun Vert $9a^3 y = x(4a - x)^3$

Differentiate y in terms of x using product formula
$9a^3 y' = -3x(4a - x)^2 + (4a - x)^3$

Simplify (optional)
$9a^3 y' = (4a - x)^2 [-3x + (4a - x)]$

$9a^3 y' = (4a - x)^2 (4a - 4x)$

$9a^3 y' = 4(4a - x)^2 (a - x)$

Determine the 2nd derivative
$9a^3 y'' = -4(4a - x)^2 - 8(4a - x)(a - x)$

$9a^3 y'' = -4(4a - x)[ (4a - x) + 2(a - x) ]$

$9a^3 y'' = -4(4a - x)(6a - 3x)$

$9a^3 y'' = -12(4a - x)(2a - x)$

Set y' to zero for maxima and/or minima
$0 = 4(4a - x)^2 (a - x)$

For (4a - x)2 = 0:

$x = 4a$

$9a^3 y = 4a(4a - 4a)^3$

$y = 0$

Check if Maxima or Minima.
$9a^3 y'' = -12(4a - 4a)(2a - 4a)$

$y'' = 0$   ←   inflection point (the curve is neither upward nor downward)

Hence, (4a, 0) is neither maximum nor minimum

For a - x = 0

$x = a$

$9a^3 y = a(4a - a)^3$

$y = 3a$

Check if Maxima or Minima.
$9a^3 y'' = -12(4a - a)(2a - a)$

$y'' = (-)$   ←   the curve is concave downward

Hence, (a, 0) is maximum.

Here is the graph of the curve with a = 1 Rate:
Francis June E.... Maraming Salamat po sir. Godbless you po. :)

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Francis June E.... Sir? Paano po naging (4a-4x) ? Yung sa simplify po

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Francis June E.... Okay na po pala sir hehe. Salamat po

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