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4.5 Optimization Problems (part 2)

Example

A page from a book is to have a printed area of 42 in2. The margins at top and bottom are 2 in each and the margins at left and right are 1 in each. Find the dimensions of the page to minimize total area.

A rectangular page diagram showing an inner printed area with margins. Top and bottom margins are labeled 2; side margins are 1.

printed area = 42 in2

find length and width of page w/ minimum total area.

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Diagram of the page with inner dimensions x and y, and outer dimensions labeled as x+2 and y+4.
  • x: width of printed area
  • y: length of printed area

printed area has area 42 in2

so, xy = 42 constraint

  • width of page: \( x + 2 \)
  • length of page: \( y + 4 \)

area of page: A = (x + 2)(y + 4)

objective

goal: minimize A

objective has two variables: use constraint to eliminate one

\[ xy = 42 \rightarrow y = \frac{42}{x} \quad \text{sub into A} \]

\(A = (x + 2) \left( \frac{42}{x} + 4 \right)\)

next steps: find critical numbers, check for max/min

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Example: Optimization Problem

A piece of cardboard is 6 in by 6 in. Small squares are removed from the corners and the flaps are folded up to make a box. Find the volume of the largest possible such box.

Square cardboard with side length 6 and corner squares of side x marked for removal.
Cross-shaped cardboard after corners are removed, with central dimensions labeled 6-2x.
3D box with height x and base side lengths 6-2x.

Maximize this:

volume: \( V(x) = (6 - 2x)^2(x) \)

\( 0 \le x \le 3 \)

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Next Steps

  • Find \( V' \)
  • Find critical numbers (\( V' = 0 \))
  • Then compare \( V \) at critical numbers and end points.
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Optimization Example

Example: Find the volume of the largest possible right circular cylinder that can be inscribed in a cone of height 4 and radius 3.

A cone of height 4 and base radius 3 with several inscribed cylinders of varying heights and radii shown inside.

largest possible cylinder → maximize the volume

A standalone cylinder with radius r and height h labeled.

volume = \[ V = \pi r^2 h \]

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Cut the cone/cylinder in half and look at cross section on xy axes

A 2D cross-section of the cone as a triangle on a coordinate plane with an inscribed rectangle representing the cylinder.
  • x: half width of rectangle (also radius of cylinder)
  • y: height of rectangle (also height of cylinder)

Corner of rectangle \( (x, y) \) must be on a line through \( (3, 0) \) and \( (0, 4) \)

equation: \[ y = -\frac{4}{3}x + 4 \]

volume of cylinder: \[ V = \pi (x)^2 (y) \]

max this → \[ V = \pi x^2 \left( -\frac{4}{3}x + 4 \right) \]

\[ 0 \le x \le 3 \]

Next steps:

  • find critical numbers
  • compare \( V \) at critical numbers and end points
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Example: Optimization Problem

A dog can run at 5 m/s and swim at 1 m/s.

A dog runs along the beach and jumps into the water and swims in a straight line to reach a ball floating in the water.

Where should the dog jump into the water to minimize the time to reach the ball?

  • A: start
  • B: the ball
  • D: the point where the dog jumps in
  • x: distance on land
  • y: distance in water
Geometry diagram of the dog's path from point A to D on land, then D to B in water.
Right triangle representing the swimming path with legs 5 and 10-x, and hypotenuse y.
\[ y = \sqrt{5^2 + (10-x)^2} \]

Note: 10 m and 5 m are given distances.

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\[ \text{time} = \frac{\text{distance}}{\text{speed}} \]

Time on land: \[ T_L = \frac{x}{5} \]

Time spent swimming: \[ T_W = \frac{\sqrt{5^2 + (10-x)^2}}{1} = \sqrt{25 + (10-x)^2} = (x^2 - 20x + 125)^{1/2} \]

Total Time Function

Total time = time on land + time in water

\[ T = \frac{x}{5} + (x^2 - 20x + 125)^{1/2} \]

minimize this

0 ≤ x ≤ 10

↑ Swim entire way (when x=0)

↑ Run until even w/ ball then jumps in (when x=10)

Next Steps

  • Find critical numbers
  • Compare T at critical numbers and at end points