Minima Maxima: 9a³y=x(4a-x)³

Submitted by Francis June E… on

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

Tags
Minima maxima

Jhun Vert Thu, 04/09/2020 - 17:13

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
9a2y=a(4a+a)3

9a2y=125a4

y=125a29   ←   not equal to 3a
 

Kindly check the given.
 

Jhun Vert Fri, 04/10/2020 - 22:02

9a3y=x(4a−x)3
 

Differentiate y in terms of x using product formula
9a3y′=−3x(4a−x)2+(4a−x)3
 

Simplify (optional)
9a3y′=(4a−x)2[−3x+(4a−x)]

9a3y′=(4a−x)2(4a−4x)

9a3y′=4(4a−x)2(a−x)
 

Determine the 2nd derivative
9a3y″=−4(4a−x)2−8(4a−x)(a−x)

9a3y″=−4(4a−x)[(4a−x)+2(a−x)]

9a3y″=−4(4a−x)(6a−3x)

9a3y″=−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
 

9a3y=4a(4a−4a)3

y=0
 

Check if Maxima or Minima.
9a3y″=−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
 

9a3y=a(4a−a)3

y=3a
 

Check if Maxima or Minima.
9a3y″=−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
 

screenshot_2020-04-10_14.23.04.png