WorkbenchDrag c and count where the line meets the circle. For which c is there only one point? Then drag P around the circle to where x+2y is largest.
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What this move does
Normally n unknowns need n equations. But the Cauchy inequality gives you one extra piece of information:
(a1x1+⋯+anxn)2≤(a12+⋯+an2)(x12+⋯+xn2).
If the given conditions make the two sides exactly equal, the equality condition is a whole chain of equations, a1x1=a2x2=⋯=anxn.
In the system, spot a “sum of products” and a “sum of squares”.
Estimate with Cauchy and notice that the given value is exactly equal to (or beyond) the bound Cauchy gives.
Equal: write (x1,…,xn)=k(a1,…,an) and substitute to find k; the solution is unique. Beyond: there is no solution.
Finding extrema is the same thing seen from the other side: when the sum of squares is fixed, the maximum of the sum of products is the Cauchy bound, and the point where it is reached is where equality holds.
Predict first
The system x2+y2=5, x+2y=c has exactly one real solution. How many such values of c are there?
Choose an answer first; then you can see the result.
✓You predicted it!✗Not what you expected — which is exactly why it is worth a pause.
Cauchy gives ∣x+2y∣≤1+4⋅5=5. For ∣c∣<5 the line cuts the circle and there are two solutions; for ∣c∣>5 there are none. Only for c=5 and c=−5 is the line tangent to the circle, with a unique solution: (1,2) and (−1,−2) respectively.
Work through one
Example: solve x² + y² = 5, x + 2y = 5
Step 1 / 5
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First, a general c
Two equations, two unknowns: it looks like a job for elimination. Start with c=3: the line cuts the circle in two points, so the system has two solutions.
x2+y2=5,x+2y=3
The bound from Cauchy
Push c up to the bound
Equality is forced
The same picture as an extremum problem
With three unknowns you cannot draw the picture, but the reasoning is word for word the same. For example, x2+y2+z2=9, 2x−y+2z=9:
(2x−y+2z)2≤(4+1+4)(x2+y2+z2)=81,
and the left side is exactly 81. So (x,y,z)=k(2,−1,2); substituting gives 9k=9, and the unique solution is (2,−1,2). Two equations, three unknowns, and yet a unique solution: the second equation has pushed Cauchy into a corner.
When it fails
Break it
Counterexample lab: remove one condition
Conclusion: the system has a unique solution← still true?
c=5 is exactly the bound: the line is tangent to the circle and there is a single intersection point.
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Change c to 4. The inequality (x+2y)2≤25 still holds, but 16<25: equality does not hold, so the equality condition tells you nothing.
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The line cuts the circle in two points and the system has two solutions, so you have to eliminate the honest way (x=4−2y gives 5y2−16y+11=0, so y=1 or 511). Cauchy only solves systems when equality is forced.
Practice
Exercise 1Basic
Real numbers x,y,z satisfy x2+y2+z2=9 and 2x−y+2z=9. Find x+y+z.
Hint · direction
Two equations, three unknowns. First use Cauchy to estimate how large 2x−y+2z can be.
Hint · method
(2x−y+2z)2≤(22+12+22)(x2+y2+z2).
Hint · first step
The right side =9×9=81, and the left side is exactly 81, so equality holds: (x,y,z)=k(2,−1,2).
Full solution
Substituting into 2x−y+2z=9k=9 gives k=1, so (x,y,z)=(2,−1,2); check: 4+1+4=9. Hence x+y+z=3.
Exercise 2Basic
Real numbers x,y satisfy x2+y2=4. Find the maximum of 3x+4y.
Hint · direction
The sum of squares is fixed; you want the maximum of a sum of products.
Hint · method
(3x+4y)2≤(9+16)(x2+y2).
Hint · first step
The right side =25×4=100, so 3x+4y≤10.
Full solution
Equality needs (x,y)=k(3,4) with k>0; from 25k2=4 we get k=52, so (x,y)=(56,58), and then 3x+4y=518+32=10. The maximum is 10.
Exercise 3Advanced
Real numbers a,b,c satisfy a+b+c=5 and a2+b2+c2=9. Find the maximum of c.
Hint · direction
Treat c as a parameter: a,b satisfy a+b=5−c and a2+b2=9−c2. What is needed for such a,b to exist?
Hint · method
Apply Cauchy to a,b: (a+b)2≤2(a2+b2).
Hint · first step
(5−c)2≤2(9−c2), which simplifies to 3c2−10c+7≤0.
Full solution
(3c−7)(c−1)≤0, so 1≤c≤37. At c=37 Cauchy is an equality and a=b=34. Check: 34+34+37=5 and 916+916+949=9. The maximum is 37.
Exercise 4Challenge
Real numbers x,y,z satisfy x+2y+2z=8. Find the minimum of (x−1)2+(y+1)2+z2.
Hint · direction
The squares are not of x,y,z themselves. Substitute so that it becomes a “plain” sum of squares.
Hint · method
Let X=x−1, Y=y+1, Z=z, and rewrite the condition in terms of X,Y,Z.
Hint · first step
x+2y+2z=(X+1)+2(Y−1)+2Z=8, i.e. X+2Y+2Z=9.
Full solution
81=(X+2Y+2Z)2≤9(X2+Y2+Z2), so the expression is ≥9. Equality needs (X,Y,Z)=k(1,2,2) with 9k=9, k=1, i.e. (x,y,z)=(2,1,2). Check: 2+2+4=8 and 1+4+4=9. The minimum is 9. Geometrically, it is the squared distance from the point (1,−1,0) to the plane x+2y+2z=8.
Checking proportionality for equality but forgetting the sign
✗
The wrong step: x2+y2=5, x+2y=−5. From the equality condition 1x=2y, take x=1,y=2.
What went wrong? Think first, then open
DiagnosisSubstituting (1,2) into the second equation gives 5, not −5. In the proportionality (x,y)=k(1,2) the factor k can be positive or negative; x+2y=5k=−5 needs k=−1.
✓
The right step: Set (x,y)=k(1,2); substituting into x+2y=−5 gives k=−1, so the solution is (−1,−2). Then check x2+y2=5.
“Write the equality condition as $(x,y)=k(a,b)$, then let the equations decide $k$”
Cauchy for Systems and Extrema ← → step · F full screen · Esc exit