Calculus

Optimization

Finding minima and maxima

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010203040MAX A=25.0x=5.00
Perimeter P = 20
x = 5.00
A(x)=x(P2−x)=x(10−x)A(x)=x\left(\frac{P}{2}-x\right)=x(10-x)
A′(x)=10−2x=0⇒x∗=5,Amax⁡=25.0A'(x)=10-2x=0 \Rightarrow x^*=5, A_{\max}=25.0

Classic optimization: maximize area of a rectangle with fixed perimeter P. Setting A'(x)=0 gives the maximum. Solution: a square (x=P/4).