xy^3 = 64.
find minimum value of x+12y.
i tried the method as above, i get x= 4th root of (64/27)….pls advice
Kingsly, here you would have to say since we want y^3, let’s split 12y as 4y + 4y + 4y. So we need x = 4y so 4y * y^3 = 64 so y = 2 and x = 8 and so x + 12y = 32 I would guess…
For the last problem, what will be the maximum value of a+b+c when a^2*b*c^4 = 1024. Please explain ?
There won’t be a maximum limit, Abhishek. If we make a and b very small (close to zero), then c will become very large (tending to infinity) and so the total will tend to infinity as well. Remember that we are talking positive real numbers here and not integers.
It was helpful. Thank you
If 2x+y =8, maximum value of x^4 * y^3, Solved it by taking 2x/4 =y/3 which implies, taking value of x =2 and y =3 so maximum is 2^4 * 3^3 but answer is given to be as 2^18 * 3^3. the answer was explained in A.M>= G.M approach. could you please explain the flaw in my logic.
Neither answer seems correct. If we don’t have a condition on x and y, then it can go to infinity. If we restrict x and y to be positive, even then the max will come when x = 16/7 and y = 24/7. You have assumed x and y to be natural numbers I guess, but I see no reason for that in the question.
Do double-check the question; the answer 2^18*3^3 would come if the question was 2x+y = 28. Perhaps you have read it incorrectly or else there is a typo in the question….
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