It is not a coincidence! On patterns in some Calculus optimization - PowerPoint PPT Presentation

It is not a coincidence! On patterns in some Calculus optimization problems. Maria Nogin California State University, Fresno mnogin@csufresno.edu

Outline 1 The basics Optimizing rectangle Optimizing rectangular box 2 The rectangular field problem Problem Observation Why? 3 The can problem Problem Observation Why?

Optimizing rectangle Out of all rectangles with a given perimeter, which one has the greatest area? s−h h h s s−h s+h

Optimizing rectangular box Out of all rectangular boxes with a given volume, which one has the smallest surface area?

The rectangular field roblem A farmer wants to fence off a rectangular field and divide it into 3 pens with fence parallel to one pair of sides. He has a total 2400 ft of fencing. What are the dimensions of the field has the largest possible area? y = 2400 − 4 x = 1200 − 2 x 2 Area( x ) = 1200 x − 2 x 2 x Area ′ ( x ) = 1200 − 4 x 2 = 0 x = 300 is an absolute maximum y y = 600 the total length of vertical pieces: 1200 ft Observation: the total length of horizontal pieces: 1200 ft These are equal!

Why? Functional explanation x y Let L be the total length of the vertical pieces. 2400 − L is the total length of the horizontal pieces. y = 2400 − L 4 · 2400 − L x = L Area( L ) = L 4 , , 2 2 Area L 0 1200 2400

Why? Geometric explanation x x x y y

Why? Geometric explanation x/ 2 x x x y y

The can problem A cylindrical can has to have volume 1000cm 3 . Find the dimensions of the can that minimize the amount of material used (i.e. minimize the surface area). h = 1000 π r 2 SA ( r ) = 2 π r 2 + 2000 h r SA ′ ( r ) = 4 π r − 2000 = 0 r 2 � 500 r r = 3 is an absolute minimum π � 500 h = 2 3 π � 500 d = 2 3 Observation: cm d = h! π

Why? h h r r V can = A circle h V cube = A square h V cube = A square A circle V can = 4 π V can SA can = 2 A circle + P circle h SA cube = 2 A square + P square h P square P circle = A square Question: is A circle ? 2 π r = 4 r 2 8 r Yes! Answer: π r 2

Is that a coincidence? P square P circle = A square Why ? A circle Equivalently: A square A hexagon A circle = = P circle P square P hexagon π r 2 4 r 2 ? = = 2 π r 8 r ? The denominator is the derivative of the numerator! ∆ r ∆ r ∆ r r r r ar 2 br 2 cr 2 = = 2 ar 2 br 2 cr

Other boxes h h h r r r Optimal shape: h = 2 r

Thank you!