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If ‘a’ and ‘b’ are two positive real numbers and a3b4 = 32768 then the minimum possible value of (3a + b) is ___________ .

Question

If ‘a’ and ‘b’ are two positive real numbers and a3b4=32768 a^3b^4 = 32768 then the minimum possible value of (3a+b) (3a + b) is ___________ .

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Solution

This problem can be solved using the AM-GM inequality (Arithmetic Mean-Geometric Mean Inequality). The AM-GM inequality states that the arithmetic mean of non-negative real numbers is greater than or equal to the geometric mean.

Here are the steps to solve the problem:

  1. Given that a^3 * b^4 = 32768, we can take the cube root of both sides to get a * b^(4/3) = 32.

  2. We want to minimize the value of 3a + b. We can rewrite this as 3a + 3*(b/3).

  3. According to the AM-GM inequality, the arithmetic mean is greater than or equal to the geometric mean. So, (3a + 3*(b/3))/2 >= sqrt[3a * 3*(b/3)].

  4. Simplifying the right side, we get sqrt[9ab] >= (3a + b)/2.

  5. Multiply both sides by 2, we get 2sqrt[9ab] >= 3a + b.

  6. We know that a * b^(4/3) = 32, so 9ab = 9 * 32^(3/4) = 144.

  7. Substituting this into the inequality, we get 2sqrt[144] >= 3a + b.

  8. Simplifying, we get 24 >= 3a + b.

So, the minimum possible value of (3a + b) is 24.

This problem has been solved

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