Question

If $y^2(y^2-6)+x^2-8x+24=0$ and the minimum value of $x^2+y^4$ is $m$ and maximum value is $M$, then find the value of $\frac{M}{6m}$

Answer 1

$\frac{3}{8}$

Answer 2

Solution:

We start with the given equation:

$$y^2(y^2-6)+x^2-8x+24=0.$$

Let $u = y^2$; then the equation becomes:

$$u(u-6) + x^2-8x+24 = 0 \quad \Longrightarrow \quad u^2 - 6u + x^2-8x+24 = 0.$$

Complete the square in $x$:

$$x^2-8x = (x-4)^2 - 16,$$

which gives:

$$u^2-6u+(x-4)^2 - 16+24=0 \quad \Longrightarrow \quad u^2-6u+(x-4)^2+8=0.$$

Next, complete the square in $u$:

$$u^2-6u = (u-3)^2-9.$$

Thus, we have:

$$(u-3)^2-9+(x-4)^2+8=0 \quad \Longrightarrow \quad (u-3)^2+(x-4)^2=1.$$

This represents a circle centered at $(4, 3)$ in the $(x, u)$-plane with radius 1.

We need to optimize:

$$x^2+y^4 = x^2+u^2.$$

Shift coordinates by setting:

$$X = x-4, \quad V = u-3.$$

Then the constraint becomes:

$$X^2+V^2=1,$$

and

$$x = X+4,\quad u = V+3.$$

So the objective function is:

$$x^2+u^2 = (X+4)^2+(V+3)^2 = X^2+8X+16 + V^2+6V+9.$$

Since $X^2+V^2=1$, we obtain:

$$x^2+u^2 = 1+8X+6V+25 = 26+8X+6V.$$

Now, to maximize or minimize $26+8X+6V$ subject to $X^2+V^2=1$, note that the extreme values occur when the linear term $8X+6V$ is maximized or minimized, which is a dot-product with the vector $(8,6)$. The maximum value is:

$$\max(8X+6V) = \sqrt{8^2+6^2} = 10,$$

and the minimum is $-10$. Thus:

$$\text{Maximum} = 26+10 = 36,\quad \text{Minimum} = 26-10 = 16.$$

Given $m = 16$ and $M = 36$, the desired value is:

$$\frac{M}{6m} = \frac{36}{6\times 16} = \frac{36}{96} = \frac{3}{8}.$$

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Final Answer: $\frac{3}{8}$

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Core Explanation:

1. Substitute $u=y^2$ and complete the squares to transform the equation into a circle: $(x-4)^2+(u-3)^2=1$.
2. Change variables: $X = x-4$ and $V = u-3$.
3. Express $x^2+y^4 = (X+4)^2+(V+3)^2$ which simplifies to $26+8X+6V$.
4. The extrema on the unit circle are obtained by dot product leading to values 36 (max) and 16 (min), thereby $\frac{36}{6\times 16}=\frac{3}{8}$.