Question

Let positive real numbers x and y be such that 3x + 4y = 14. the maximum value of x3y4 is

Answer 1

Breaking down 3x into 3x/2 and 3x/2, and 4y into 4y/3, 4y/3, and 4y/3, and applying the inequality gives

Arithmetic Mean of (3x/2, 3x/2, 4y/3, 4y/3, 4y/3) >= Geometric Mean of (3x/2, 3x/2, 4y/3, 4y/3, 4y/3)

=> (3x/2+3x/2+4y/3+4y/3+4y/3)/5 >= (3/2*3/2*4/3*4/3*4/3*6)^(1/5)

=> (3x + 4y)/5 >= (2^5)^(1/5)

=> (3x + 4y)/5 >= 2

=> (3x + 4y) >=10
=> Minimum Value of (3x + 4y) = 10

Answer 2

Answer:

128

Step-by-step explanation:

GIven both x and y are positive and such that 3x+4y=14,

x³y⁴ = x³ ( 14 - 3x / 4 )⁴ = f(x)

Now for the maximum f ' (x) = 0

3x²  ( 14 - 3x / 4 )⁴ = 3x³ ( 14 - 3x / 4 )³

4x = 14 − 3x

x = 2

max ( f(x) ) = 128