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What is the largest topless trapezoid that can be made by folding up - - PowerPoint PPT Presentation

Oct 12, 2022 103 likes •528 views

Read the problem Formul Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema What is the largest topless trapezoid that can be made by folding up the ends of a 30 in length of wire? Read the problem

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SLIDE 1

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

What is the largest ‘topless’ trapezoid that can be made by folding up the ends of a 30 in length of wire?

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SLIDE 2

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

What is the largest ‘topless’ trapezoid that can be made by folding up the ends of a 30 in length of wire? Trying to maximise area

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SLIDE 3

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

What is the largest ‘topless’ trapezoid that can be made by folding up the ends of a 30 in length of wire? Trying to maximise area Why not like this?

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SLIDE 4

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Area of a trapezoid: A = 1 2(t + b)h t and b are length of top and bottom, and h is the height (in inches)

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SLIDE 5

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h

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SLIDE 6

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b + 2s = 30 ⇒ b = 30 − 2s By Pythagorean theorem, s2 = h2 + e2 ⇒ e =

  • s2 − h2
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SLIDE 7

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b + 2s = 30 ⇒ b = 30 − 2s By Pythagorean theorem, s2 = h2 + e2 ⇒ e =

  • s2 − h2

and t = b + 2e = 30 − 2s + 2

  • s2 − h2.
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SLIDE 8

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b = 30 − 2s and t = 30 − 2s + 2

  • s2 − h2
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SLIDE 9

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b = 30 − 2s and t = 30 − 2s + 2

  • s2 − h2

New formula: A = 1 2(t + b)h

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SLIDE 10

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b = 30 − 2s and t = 30 − 2s + 2

  • s2 − h2

New formula: A = 1 2

  • (30 − 2s + 2
  • s2 − h2) + (30 − 2s)
  • h
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SLIDE 11

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Three input variables, t, b, h, plus two more, s and e: b = 30 − 2s and t = 30 − 2s + 2

  • s2 − h2

New formula: A = 1 2

  • (30 − 2s + 2
  • s2 − h2) + (30 − 2s)
  • h

= (30 − 2s +

  • s2 − h2)h
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SLIDE 12

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

t, b, h, s, and e are lengths, so all are ≥ 0

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SLIDE 13

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

t, b, h, s, and e are lengths, so all are ≥ 0 b = 30 − 2s ≥ 0 ⇒ s ≤ 15

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SLIDE 14

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

t, b, h, s, and e are lengths, so all are ≥ 0 b = 30 − 2s ≥ 0 ⇒ s ≤ 15 s2 = h2 + e2 ⇒ s ≥ h

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SLIDE 15

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

t, b, h, s, and e are lengths, so all are ≥ 0 b = 30 − 2s ≥ 0 ⇒ s ≤ 15 s2 = h2 + e2 ⇒ s ≥ h Domain is (s, h) with 0 ≤ s ≤ 15 and 0 ≤ h ≤ s

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SLIDE 16

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

t, b, h, s, and e are lengths, so all are ≥ 0 b = 30 − 2s ≥ 0 ⇒ s ≤ 15 s2 = h2 + e2 ⇒ s ≥ h Domain is (s, h) with 0 ≤ s ≤ 15 and 0 ≤ h ≤ s

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SLIDE 17

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Interior of the domain is where all inequalities are strict: 0 < s < 15 and 0 < h < s.

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SLIDE 18

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Interior of the domain is where all inequalities are strict: 0 < s < 15 and 0 < h < s. Find where ∇A = ∂A ∂s , ∂A ∂h

  • =

(or ∇A is undefined)

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SLIDE 19

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Interior of the domain is where all inequalities are strict: 0 < s < 15 and 0 < h < s. Find where ∇A = ∂A ∂s , ∂A ∂h

  • =

(or ∇A is undefined) Be careful! There are solutions not in the interior of the domain

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SLIDE 20

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and 0 ≤ h ≤ s

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SLIDE 21

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and 0 ≤ h ≤ 0

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SLIDE 22

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and h = 0

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SLIDE 23

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and h = 0 s = 15 and 0 ≤ h ≤ s

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SLIDE 24

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and h = 0 s = 15 and 0 ≤ h ≤ 15

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SLIDE 25

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and h = 0 s = 15 and 0 ≤ h ≤ 15 0 ≤ s ≤ 15 and h = 0

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SLIDE 26

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Boundary is where we have equality: s = 0 and h = 0 s = 15 and 0 ≤ h ≤ 15 0 ≤ s ≤ 15 and h = 0 0 ≤ s ≤ 15 and h = s

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SLIDE 27

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 0, the only possibility is h = 0, so add (0, 0) to the list

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SLIDE 28

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15:

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SLIDE 29

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2s +

  • s2 − h2)h
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SLIDE 30

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2(15) +

  • (15)2 − h2)h
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SLIDE 31

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2(15) +

  • (15)2 − h2)h

=

  • 225 − h2 · h.
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SLIDE 32

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2(15) +

  • (15)2 − h2)h

=

  • 225 − h2 · h.

Domain is 0 ≤ h ≤ s.

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SLIDE 33

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2(15) +

  • (15)2 − h2)h

=

  • 225 − h2 · h.

Domain is 0 ≤ h ≤ 15. One critical point (s, h) = (15, h) with 0 < h < 15 (where?)

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SLIDE 34

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Treat each case separately When s = 15: A = (30 − 2(15) +

  • (15)2 − h2)h

=

  • 225 − h2 · h.

Domain is 0 ≤ h ≤ 15. One critical point (s, h) = (15, h) with 0 < h < 15 (where?) Endpoints (s, h) = (15, 0) and (s, h) = (15, 15)

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SLIDE 35

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

List so far: Interior critical point

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SLIDE 36

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

List so far: Interior critical point (0, 0)

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SLIDE 37

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

List so far: Interior critical point (0, 0) (15, 0) and (15, 15), and another critical point where s = 15

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SLIDE 38

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

List so far: Interior critical point (0, 0) (15, 0) and (15, 15), and another critical point where s = 15 . . .

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SLIDE 39

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Plug all points (s, h) into A: s h A 15 15 15 . . . . . . . . .

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SLIDE 40

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Plug all points (s, h) into A: s h A 15 15 15 . . . . . . . . . The biggest are the absolute maxima

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SLIDE 41

Read the problem Formulæ Cut down independent variables Domain Calculus (interior) Calculus (boundary) Extrema

Plug all points (s, h) into A: s h A 15 15 15 . . . . . . . . . The biggest are the absolute maxima; no need for derivative tests