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

# Minima Maxima: 9a³y=x(4a-x)³

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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=aand solve fory$9a^2 y = a(4a + a)^3$

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

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

aKindly check the given.

Ay sorry po.

Eto po sir 9a²y=x(4a-x)³

The same error. You can evaluate if

x=a,y= 3a^{2}not 3a. Either the given is wrong or the answer key is wrong.Kinuha ko kasi sa libro ni Love and Rainville yang example sir.

i-check ko. Anong page?

62-64 sir

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

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

Differentiate

yin terms ofxusing 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 2

^{nd}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 (4

a-x)^{2}= 0:$9a^3 y = 4a(4a - 4a)^3$

$y = 0$

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, (4

a, 0) is neither maximum nor minimumFor

a-x= 0$9a^3 y = a(4a - a)^3$

$y = 3a$

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= 1Maraming Salamat po sir. Godbless you po. :)

Sir? Paano po naging (4a-4x) ? Yung sa simplify po

Okay na po pala sir hehe. Salamat po