Optimization by Equalization

Idea · Intermediate

Algebra

a,b≥0a,b\ge 0
Fixed quantityQuantity optimizedMinimumOccurs whenMaximumOccurs when
a+b=Sa+b=Sabab00a,ba,b are as far apart as possibleS24\dfrac{S^2}{4}a,ba,b are as close as possible
a+b=Sa+b=Sa2+b2a^2+b^2S22\dfrac{S^2}{2}a,ba,b are as close as possibleS2S^2a,ba,b are as far apart as possible
ab=P>0ab=P>0a+ba+b2P2\sqrt Pa,ba,b are as close as possibleNo maximuma,ba,b become arbitrarily far apart
ab=P>0ab=P>0a2+b2a^2+b^22P2Pa,ba,b are as close as possibleNo maximuma,ba,b become arbitrarily far apart
a2+b2=Qa^2+b^2=Qa+ba+bQ\sqrt Qa,ba,b are as far apart as possible2Q\sqrt{2Q}a,ba,b are as close as possible
a2+b2=Qa^2+b^2=Qabab00a,ba,b are as far apart as possibleQ2\dfrac Q2a,ba,b are as close as possible

Key Idea

For optimization problems involving two nonnegative numbers:

Look at what happens when a and b become closer together or farther apart\boxed{\text{Look at what happens when }a\text{ and }b\text{ become closer together or farther apart}}

If a and b are real numbers, as close as possible often means

a=ba=b

If a and b are integers, equality may be impossible. In that case, choose them as close as possible.

For example, if

a+b=15,a,b∈Z≥0a+b=15,\qquad a,b\in\mathbb Z_{\ge0}

the closest possible pair is

7,8\boxed{7,8}

not 7.5, 7.5.