1thanks sir.
5 Answers
62The trick for this one is that
3x/2+3x/2+4y/3+4y/3+4y/3=1
Now apply am GM To get...
\frac{3x/2+3x/2+4y/3+4y/3+4y/3}{5}\ge \left( 3x/2\times3x/2\times4y/3\times4y/3\times4y/3\right)^{1/5}
Sovle the rest..
39As x and y are not given to be strictly positive, we can't use axioms of algebra. Instead we shall use calculus.
y = (1 - 3x)4
Apply this substitution to our expression.
Z = x²(1 - 3x)4
=> Z = x² - 3x34
This is a function of x, so we write it as f(x).
f'(x) = 2x - 9x²4
Put f'(x) = 0.
x(2 - 9x) = 0
This implies that x = 0, 2/9.
f'(x) changes sign across 0 and 2/9.
Now f"(x) = 2 - 18x4
f"(0) > 0 and hence x = 0 is a point of minima.
So x = 2/9 is a point of maxima.
So the maximum value is f(2/9) = (29)² - 3(29)34
We could use AM-GM though.
Okay nishant bhaiya has solved it that way :)
62sorry vardaan I overlooked the thing that x and y were not supposedly +ve.
btw if they are allowed to be -ve then the answer will simply be infinite..
1where am i wrong
x2y3 = x x y y y
x repeated 2 times and y repeated 3 times
given 3x + 4y =1 → 2(3x2) + 3(4y3) = 1
→ 3x2 + 3x2 + 4y3 + 4y3 + 4y3 = 1
greatest value of
3x2 . 3x2 . 4y3 . 4y3 . 4y3 is \left(\frac{1}{5} \right)^{5}
hence greatest value of 32432233.x2y3 = 1
hence greatest value of x2y3 = 22333243 = 0.1875