Question

If a,b,c are positive real numbers, such that minimum value of E = a⁴ + 2b⁴ + 4c⁴ – 8abc is 'm' and a₀,b₀,c₀ are respective value of a,b,c for which E is minimum, then -

• A. m = 2
• B. m = -2
• C. a₀b₀c₀ = 1
• D. a₀b₀c₀ = 4

Answer 1

Answer: (B), (C)

B, C

Answer 2

To find the minimum value of the expression $E = a^4 + 2b^4 + 4c^4 - 8abc$ for positive real numbers $a,b,c$, we use calculus by finding the partial derivatives with respect to $a, b, c$ and setting them to zero.

Let $E(a,b,c) = a^4 + 2b^4 + 4c^4 - 8abc$. The partial derivatives are: $\frac{\partial E}{\partial a} = 4a^3 - 8bc$ $\frac{\partial E}{\partial b} = 8b^3 - 8ac$ $\frac{\partial E}{\partial c} = 16c^3 - 8ab$

At the point of minimum $(a_0, b_0, c_0)$, these partial derivatives must be zero:

1. $4a_0^3 - 8b_0c_0 = 0 \implies a_0^3 = 2b_0c_0$
2. $8b_0^3 - 8a_0c_0 = 0 \implies b_0^3 = a_0c_0$
3. $16c_0^3 - 8a_0b_0 = 0 \implies 2c_0^3 = a_0b_0$

From equation (2), $c_0 = \frac{b_0^3}{a_0}$. Substitute this into equation (1): $a_0^3 = 2b_0 \left(\frac{b_0^3}{a_0}\right) = \frac{2b_0^4}{a_0}$ $a_0^4 = 2b_0^4$ (Equation I)

Substitute $c_0 = \frac{b_0^3}{a_0}$ into equation (3): $2\left(\frac{b_0^3}{a_0}\right)^3 = a_0b_0$ $\frac{2b_0^9}{a_0^3} = a_0b_0$ $2b_0^9 = a_0^4b_0$ Since $b_0 > 0$, we can divide by $b_0$: $2b_0^8 = a_0^4$ (Equation II)

Now we equate Equation I and Equation II: $2b_0^4 = 2b_0^8$ $b_0^8 - b_0^4 = 0$ $b_0^4(b_0^4 - 1) = 0$ Since $b_0$ is a positive real number, $b_0^4 \neq 0$. Therefore, $b_0^4 - 1 = 0$, which implies $b_0^4 = 1$. As $b_0 > 0$, we get $b_0 = 1$.

Now we find $a_0$ and $c_0$: From Equation I: $a_0^4 = 2b_0^4 = 2(1)^4 = 2$. Since $a_0 > 0$, $a_0 = 2^{1/4}$.

From equation (2): $b_0^3 = a_0c_0$. $1^3 = (2^{1/4})c_0 \implies c_0 = \frac{1}{2^{1/4}} = 2^{-1/4}$.

So, the values for which $E$ is minimum are $a_0 = 2^{1/4}$, $b_0 = 1$, and $c_0 = 2^{-1/4}$.

Now, let's calculate the minimum value $m$: $m = E(a_0, b_0, c_0) = a_0^4 + 2b_0^4 + 4c_0^4 - 8a_0b_0c_0$. $a_0^4 = (2^{1/4})^4 = 2$. $b_0^4 = 1^4 = 1$, so $2b_0^4 = 2(1) = 2$. $c_0^4 = (2^{-1/4})^4 = 2^{-1} = 1/2$, so $4c_0^4 = 4(1/2) = 2$. $a_0b_0c_0 = (2^{1/4})(1)(2^{-1/4}) = 2^{1/4 - 1/4} = 2^0 = 1$. So, $8a_0b_0c_0 = 8(1) = 8$.

$m = 2 + 2 + 2 - 8 = 6 - 8 = -2$.

Checking the options: (A) m = 2: False. (B) m = -2: True. (C) $a_0b_0c_0 = 1$: True. (D) $a_0b_0c_0 = 4$: False.

The minimum value of $E$ is $m=-2$, and this occurs at $a_0=2^{1/4}, b_0=1, c_0=2^{-1/4}$. For these values, $a_0b_0c_0 = 1$. Therefore, options (B) and (C) are correct.