Max & Min Problems (Leaving Cert Mathematics): Flashcards

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Max & Min Problems
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Method to find optimization max/min

Derive, set to 0, solve

Derivative condition for local max/min

Set derivative equal to zero

Derivative of P(x)=−x2+10xP(x) = -x^2 + 10x

P′(x)=−2x+10P'(x) = -2x + 10

Optimal cakes for max profit (P(x)=−x2+10xP(x) = -x^2 + 10x)

5 cakes

Max profit value (5 cakes, P(x)=−x2+10xP(x) = -x^2 + 10x)

25 euros

Rectangle perimeter equation (100m total)

2l+2w=1002l + 2w = 100

ww in terms of ll (perimeter 100m)

w=50−lw = 50 - l

Area expression for rectangle (given w=50−lw = 50 - l)

A=50l−l2A = 50l - l^2

Derivative of area A=50l−l2A = 50l - l^2

A′=50−2lA' = 50 - 2l

Optimal ll and ww (100m perimeter rectangle)

l=25l = 25m, w=25w = 25m

Cylinder volume equation (1 litre = 1000 cm³)

πr2h=1000\pi r^2 h = 1000

Cylinder hh in terms of rr (volume 1000 cm³)

h=1000πr2h = \frac{1000}{\pi r^2}

Cost function for cylinder (radius rr)

C(r)=2πr2+2000rC(r) = 2\pi r^2 + \frac{2000}{r}

Meaning when C′′(r)>0C''(r) > 0 at critical point

Local minimum

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