Engineering Mathematics · Maxima & Minima
Maximum and Minimum of \(f(x,y)=4x^2-2xy+y^2-8x+2y+1\)
Problem 6
Investigate the maximum and minimum values of the function \(f(x,y)=4x^{2}-2xy+y^{2}-8x+2y+1\).Step 1: First-order partial derivatives
\[ f_x = \frac{\partial f}{\partial x} = 8x-2y-8 \qquad f_y = \frac{\partial f}{\partial y} = -2x+2y+2 \]Step 2: Find the stationary point
Set \(f_x=0\) and \(f_y=0\): \[ 8x-2y-8=0 \Rightarrow 4x-y=4 \Rightarrow y=4x-4 \] \[ -2x+2y+2=0 \Rightarrow -x+y=-1 \Rightarrow y=x-1 \] Equate the two values of \(y\): \(4x-4=x-1\), so \(3x=3\) and \(x=1\). Then \(y=1-1=0\).The stationary point is \((1,0)\).
Step 3: Second-order derivatives
\[ A=f_{xx}=8 \qquad B=f_{xy}=-2 \qquad C=f_{yy}=2 \]Step 4: Test the point
| \((1,0)\) | |
|---|---|
| \(A=8\) | \(8>0\) |
| \(B=-2\) | \(-2\) |
| \(C=2\) | \(2\) |
| \(AC-B^2\) | \(16-4=12>0\) |
| Result | Minimum (\(A>0\)) |
Step 5: Minimum value
\[ f(1,0)=4(1)^2-2(1)(0)+(0)^2-8(1)+2(0)+1 = 4-0+0-8+0+1 = -3 \]Answer
Minimum value = −3 at \((1,0)\).
There is no maximum value, since the only stationary point is a minimum.
There is no maximum value, since the only stationary point is a minimum.
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