Applications of Derivatives

Optimization from a Given Model

Minimize C(x)=x²-6x+11 by finding where C'(x)=2x-6=0, giving x=3, C(3)=2.

Example

Find and classify the critical point of a given objective model.

highlighted = computed this step

Step 1 — Set up model

Start with the objective model.

C(x)=x2−6x+11C(x)= x^{2}-6x+11

Step 2 — Differentiate

Differentiate the objective.

C′(x)=2x−6C'(x)= \hlmath{2x-6}

Step 3 — Critical point

Set C prime equal to 0 to get x equals 3.

2x−6=0⇒x=32x-6 = 0 \Rightarrow x= \hl{3}

Step 4 — Classify

Because C double prime is positive, this is a minimum.

C′′(x)=2>0⇒minimumC''(x)= \hlmath{2} > 0 \Rightarrow \hl{minimum}

Step 5 — Optimum value

Evaluate C at 3 to get optimum value 2.

C(3)=2C( 3 )= \hl{2}
optimization-model-given To optimize a given function: find critical points by setting C'(x)=0, then use C''(x) to classify (positive means minimum, negative means maximum), and evaluate C at the critical point to find the optimal value.