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The minimum value of \( f(x, y) = 4x^4 + 4y^4 - xy \) is \(-\frac{1}{32}\).
Solución
We wish to find the minimum value of
\[
f(x,y)=4x^4+4y^4-xy.
\]
This is achieved by finding the critical points of \(f\) and then comparing the function values.
### Step 1. Compute the Partial Derivatives
The partial derivative with respect to \(x\) is:
\[
f_x(x,y)=\frac{\partial f}{\partial x}=16x^3-y.
\]
The partial derivative with respect to \(y\) is:
\[
f_y(x,y)=\frac{\partial f}{\partial y}=16y^3-x.
\]
### Step 2. Set the Partial Derivatives Equal to Zero
We solve the system:
\[
\begin{cases}
16x^3-y=0,\\[1mm]
16y^3-x=0.
\end{cases}
\]
From the first equation, we have:
\[
y=16x^3.
\]
Substitute this into the second equation:
\[
16(16x^3)^3-x=0.
\]
### Step 3. Simplify the System
First, compute \((16x^3)^3\):
\[
(16x^3)^3=16^3x^9=4096x^9.
\]
Then the second equation becomes:
\[
16\cdot4096x^9-x=0,
\]
or
\[
65536x^9-x=0.
\]
We factor out \(x\):
\[
x(65536x^8-1)=0.
\]
### Step 4. Solve for \(x\)
This gives two cases:
1. \(x=0\).
Then using \(y=16x^3\), we obtain:
\[
y=0.
\]
So one critical point is \((0,0)\).
2. \(65536x^8-1=0\).
Solve for \(x^8\):
\[
x^8=\frac{1}{65536}.
\]
Notice that
\[
65536=2^{16},
\]
so
\[
x^8=2^{-16}.
\]
Taking the eighth root, we find:
\[
|x|=2^{-2}=\frac{1}{4}.
\]
Thus, \(x=\frac{1}{4}\) or \(x=-\frac{1}{4}\).
For \(x=\frac{1}{4}\), we compute:
\[
y=16\left(\frac{1}{4}\right)^3=16\cdot\frac{1}{64}=\frac{1}{4}.
\]
For \(x=-\frac{1}{4}\), we get:
\[
y=16\left(-\frac{1}{4}\right)^3=16\cdot\left(-\frac{1}{64}\right)=-\frac{1}{4}.
\]
Thus, the other two critical points are \(\left(\frac{1}{4},\frac{1}{4}\right)\) and \(\left(-\frac{1}{4},-\frac{1}{4}\right)\).
### Step 5. Evaluate \(f(x,y)\) at the Critical Points
1. At \((0,0)\):
\[
f(0,0)=4(0)^4+4(0)^4-0\cdot0=0.
\]
2. At \(\left(\frac{1}{4},\frac{1}{4}\right)\):
\[
f\left(\frac{1}{4},\frac{1}{4}\right)=4\left(\frac{1}{4}\right)^4+4\left(\frac{1}{4}\right)^4-\left(\frac{1}{4}\right)\left(\frac{1}{4}\right).
\]
We note that
\[
\left(\frac{1}{4}\right)^4=\frac{1}{256}.
\]
Therefore,
\[
f\left(\frac{1}{4},\frac{1}{4}\right)=4\cdot\frac{1}{256}+4\cdot\frac{1}{256}-\frac{1}{16}=\frac{4}{256}+\frac{4}{256}-\frac{1}{16}.
\]
Simplify:
\[
\frac{4}{256}=\frac{1}{64},
\]
so
\[
f\left(\frac{1}{4},\frac{1}{4}\right)=\frac{1}{64}+\frac{1}{64}-\frac{1}{16}=\frac{2}{64}-\frac{4}{64}=-\frac{2}{64}=-\frac{1}{32}.
\]
3. At \(\left(-\frac{1}{4},-\frac{1}{4}\right)\):
By symmetry, the value is the same:
\[
f\left(-\frac{1}{4},-\frac{1}{4}\right)=-\frac{1}{32}.
\]
### Step 6. Determine the Minimum Value
Among the computed values:
\[
f(0,0)=0 \quad \text{and} \quad f\left(\frac{1}{4},\frac{1}{4}\right)=f\left(-\frac{1}{4},-\frac{1}{4}\right)=-\frac{1}{32},
\]
the minimum value is:
\[
-\frac{1}{32}.
\]
Thus, the minimum value of \(f(x,y)\) is \(-\frac{1}{32}\).
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