tdec110 Lecture # 9 Optimization Overview

Transcription

1 tdec0 Lecture # 9 Optimization Overview Most "real-worl" problems are concerne wit maximizing or minimizing some quantity or entity. Te calculus is te tool tat te engineer uses to fin te BEST SOLUTIONS to tese practical problems.

2 Optimization Problem Solving Process Draw a iagram sowing only te essentials. Give te iagram symbols. Analyze te iagram relating te "Knowns" to te "Unknowns". Fin te extreme values using calculus. Confirm your finings wit a grap. "Fin te imensions tat will minimize te cost of te metal to manufacture a liter can." Example # / Part Minimum surface area results in minimal cost. But te can must be liter in volume. A = r + r V = r = 000 cm

3 Example # / Part Recall te equations: Eliminate in te equation for te surface A = r + r area. V = r 000 Ar () r 000 := + r Solve for in terms of r in te constraint equation. r () 000 r Fin te critical numbers. Tese are te values of r were, r Ar () 0 Example # / Part Differentiate A(r) an set te result equal to zero. r Ar () 4 r r Solve for r to obtain te critical number: r A(5.49) is a minimum because: A '(r) <0 r < 5.49 A '(r) >0 r > 5.49 Determine te corresponing value of : ( 5.49)

4 Example # / Part 4 Te ceapest liter can: A 55.58cm r 5.49 cm 0.89 cm A circular cyliner wit a fixe volume as te least surface area wen = r. Ar () r Example # / Part A rain gutter is to be constructe from seet metal 0 cm wie by bening it 0 cm in from bot ens. Fin te angle of tose bens tat will result in te maximum water-carrying capacity of te gutter Capacity is a maximum wen te cross-sectional area is largest. Gutter area is te sum of a rectangular an two equal triangular areas. Ar θ At θ A θ ( ) := Ar( θ) + At ( θ) A θ ( ) := 0 0 ( ) := sin( θ) ( ) 0 sin θ 0 ( ) := 00 sin( θ) cos( θ) ( ( )) + cos θ

5 Example # / Part Fin te critical numbers. Tese are te values of te angle, were: ( ) := 00 sin( θ) A θ θ A( θ) 0 ( ( )) + cos θ θ A( θ) ( ( )) ( + cos( θ) ) 00 cos θ ( ( ) ) 00 cos θ + sin θ ( cos( θ) + ) 0 ( ) ( sin( θ) ) Example # / Part Tere are two solutions to te equation: 00 ( cos( θ) ) ( cos( θ) + ) 0 But only te first one applies. A( θ) θ θ θ

6 Example # / Part A rope is strung from te tops of vertical poles an between te poles it is tie to a point on te groun. Sow tat te sortest lengt of rope occurs wen te angles tat te rope makes wit te groun are equal. Lx () P + x + S + ( x) x Lx () x x + P x S + ( x) Example # / Part Because: x Lx () x x + P x S + ( x) An, cos( θ ) x x + P cos( θ ) x S + ( x) Ten, x Lx () Set te erivative equal to zero an fin tat: cos( θ ) cos( θ ) θ θ

7 Example # 4 / Part "A man is trappe in a swamp at te location : 4, 0. Te rescue team must stay as close to im as possible so tat tey can pass a rope to im. Fin te best location of te rescue team an te minimum lengt of rope require to save im." Te best spot for te rescue team is at te point : ( x 0, y 0 ) locate somewere along te swamp bounary. Let "L" be te istance between te team an te trappe man. L x 0 + ( y 0 ) 4 Implicitely Differentiate te expression for "L" wit respect to " x 0 ". Set te result to zero an ten solve for " y 0 ". x 0 Example # 4 / Part L x 0 ( ) x 0 + ( ) ( y 0 ) y 0 4 x 0 0 x 0 + ( y 0 ) 4 y 0 x 0 x 0 4 y 0 Tis expression for " y 0 " is te slope of te x0 tangent line to te curve at te point: ( x 0, y 0 ). Te slope of te tangent line is also equal to: " x 0 ". x 0 4 x 0 y 0 Solve for " y 0 ". x 0 4 y 0 x 0

8 Example # 4 / Part But te point: ( x 0, y 0 ) lies on te parabola. y 0 x 0 x 0 ( ) 4 x 0 Tis prouces a cubic equation for " x 0 ". ( x 0 ) + x Solve te cubic grapically. f( x) x Te solution is x 0 an tus x ( 0, y 0 ), 4. Te minimum lengt, " L min ", is te istance between te two points:, 4 an 4, 0. L min Example # 5 / Part "Fin te most avantageous lengt of a lever to raise a weigt of 500 pouns if te istance of te weigt from te fulcrum is one foot an te lever weigs pouns-per-foot." Te most avantageous lever is te one tat requires te least force, "P", to lift te system. Te sum of all te moments must be zero. Let "L" be te lengt of te lever. 500 ( ) L + L P ( L) 0 Solve for "P" in terms of "L". P 500 L + L

9 Example # 5 / Part Fin " P" an set te result to zero. L L P 500 L + 0 Set te result to zero an solve for "L". L ft PL ( ) L Example # 6 / Part "A lamp is suspene above te center of a roun table of raius, "r". How ig above te table soul te lamp be place to acieve te maximum illumination at te ege of te table?" [ Assume tat te illumination, "I", is irectly proportional to te cosine of te angle of incience, " φ", of te ligt rays an inversely proportional to te square of te istance, "L", from te ligt source.] I ( ) I 0 cos φ L I 0 L I 0 ( ) + r

10 Example # 6 / Part I ( ) I 0 ( ) + r Fin " I ( )" an set te result equal to zero. I ( ) I 0 ( ) + r ( ) I 0 ( ) 0 5 ( + r ) Solve for "". ( ) + r 0 r Demonstrate wit a grap tat te ratio : is te optimal one. r Example # 6 / Part 4 I ( ) I 0 ( + r ) I r r ( I 0 r ) + r Since te factor : I 0 r varies wit te ratio: r. Let x ( ) is a constant, te illumination ix ( ) := ( x) + ( x) r an efine i(x). ix ( ) x

11 Example # 7 / Part "A cone is mae from a circular seet of raius, "R", by cutting out a sector an gluing te cut eges of te remaining piece togeter. Wat is te maximum volume attainable for te cone?" V Let "" be te eigt of te cone. ( ) R ( ) ( ) R Fin " V ( )" an set te result to zero. V ( ) R 0 Solve for "". R Example # 7 / Part Te Maximum Attainable Volume, "V max " is foun by R setting in te expression for "V()". V max R R R R 9 Demonstrate wit a grap tat te ratio: results in te cup wit te largest volume. R Since te factor: R is a constant, "V" varies as te ratio: R. Let x an efine "v(x)". R ( ) v( x) := x x V( ) R R R vx ( ) x

12 Black Slie