Minima Maxima: 9a³y=x(4a-x)³
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I think something is wrong
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.
Ay sorry po.
Ay sorry po.
Eto po sir 9a²y=x(4a-x)³
The same error. You can
The same error. You can evaluate if x = a, y = 3a2 not 3a. Either the given is wrong or the answer key is wrong.
Kinuha ko kasi sa libro ni
Kinuha ko kasi sa libro ni Love and Rainville yang example sir.
i-check ko. Anong page?
i-check ko. Anong page?
62-64 sir
62-64 sir
Okay I got it. The given is
Okay I got it. The given is actually $9a^3 y = x(4a - x)^3$
$9a^3 y = x(4a - x)^3$
$9a^3 y = x(4a - x)^3$
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:
$9a^3 y = 4a(4a - 4a)^3$
$y = 0$
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
$9a^3 y = a(4a - a)^3$
$y = 3a$
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
Maraming Salamat po sir.
Maraming Salamat po sir. Godbless you po. :)
Sir? Paano po naging (4a-4x)
Sir? Paano po naging (4a-4x) ? Yung sa simplify po
Okay na po pala sir hehe.
Okay na po pala sir hehe. Salamat po