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Optimization Problems
By Tuesday J. Johnson
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Suggested Review Topics
β’ Algebra skills reviews suggested:
β None
β’ Trigonometric skills reviews suggested:
β None
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Applications of Differentiation
Optimization Problems
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Guidelines for Solving Applied Minimum and Maximum Problems
1. Identify all given quantities and all quantities to be determined. If possible, or necessary, make a sketch.
2. Write a preliminary equation for the quantity that is to be maximized or minimized.
3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.
4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.
5. Determine the desired maximum or minimum using the Calculus techniques we have developed.
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Find two positive numbers such that the product is 185 and the sum is a minimum.
β’ Two positive numbers: let them be π₯ and π¦.
β’ The product is 185: π₯π¦ = 185.
β’ Sum is a minimum: π = π₯ + π¦.
The sum equation is our primary equation. We want to use the product equation to help
eliminate a variable. If π₯π¦ = 185, then π¦ =185
π₯.
We can now focus on π in one variable.
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Find two positive numbers such that the product is 185 and the sum is a minimum.
Sum is a minimum: π = π₯ + π¦ = π₯ +185
π₯
The minimum will occur at a critical number so we find the derivative:
π = π₯ +185
π₯= π₯ + 185π₯β1
πβ² = 1 β 185π₯β2 = 1 β185
π₯2
πβ does not exist at π₯ = 0, but this is not a positive number so letβs solve πβ = 0.
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Find two positive numbers such that the product is 185 and the sum is a minimum.
Sum is a minimum: π = π₯ + π¦ = π₯ +185
π₯
0 = 1 β185
π₯2
185
π₯2= 1
So π₯2 = 185 and π₯ = 185. We only use the positive square root based on the conditions of the problem.
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Find two positive numbers such that the product is 185 and the sum is a minimum.
Sum is a minimum: π = π₯ + π¦ = π₯ +185
π₯
First derivative: πβ² = 1 β185
π₯2
How do we know π₯ = 185 is a minimum?
Second derivative: π" = 2(185)π₯β3 =370
π₯3 and
π"( 185) is positive. By the 2nd derivative test this is a minimum.
If π₯ = 185 and π¦ =185
π₯, then π¦ = 185 also.
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Find the length and width of a rectangle that has a perimeter of 80 meters and a maximum area.
β’ π = 80 = 2πΏ + 2π or
β’ 40 = πΏ + πor
β’ 40 β π = πΏ
β’ Primary equation is area with π΄ = πΏπ = 40 β π π = 40π β π2
π΄β² = 40 β 2π
π΄β² = 0 when 0 = 40 β 2π so π = 20. The second derivative is always negative so this is a maximum.
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W
L
Find the length and width of a rectangle that has a perimeter of 80 meters and a maximum area.
β’ π = 80 = 2πΏ + 2π or
β’ 40 = πΏ + πor
β’ 40 β π = πΏ
β’ Primary equation is area with π΄ = 40π β π2 π΄β² = 40 β 2π
π΄β² = 0 so π = 20. If π = 20, then
πΏ = 40 β 20 = 20.
A rectangle of max area will always be a square. 10
W
L
Find the point on the graph of the function π π₯ = (π₯ β 1)2 that is closest to the point (β5,3).
β’ βClosest toβ tells us to find the minimum distance.
β’ Using the distance formula and the points (β5,3) and π₯, π₯ β 1 2 we have:
π = (β5 β π₯)2+(3 β π₯ β 1 2)2
This does not look fun, the derivative will be easier if we simplify the radicand to a polynomial.
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π = (β5 β π₯)2+(3 β π₯ β 1 2)2
π = π₯4 β 4π₯3 + π₯2 + 18π₯ + 29
Now we can find the derivative to find the minimum:
πβ² =1
2π₯4 β 4π₯3 + π₯2 + 18π₯ + 29 β
12(4π₯3
β 12π₯2 + 2π₯ + 18)
πβ² =4π₯3 β 12π₯2 + 2π₯ + 18
2 π₯4 β 4π₯3 + π₯2 + 18π₯ + 29
πβ² =2π₯3 β 6π₯2 + π₯ + 9
π₯4 β 4π₯3 + π₯2 + 18π₯ + 29
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Find the point on the graph of the function π π₯ = (π₯ β 1)2 that is closest to the point (β5,3).
Minimize: π = (β5 β π₯)2+(3 β π₯ β 1 2)2
With πβ² =2π₯3β6π₯2+π₯+9
π₯4β4π₯3+π₯2+18π₯+29
The derivative will be zero when the numerator is zeroβ¦using techniques from Pre-Calculus we find this to be at π₯ = β1. (The other two roots are complex).
We could use the 2nd derivative test to see if this is a minimum, or assume we are excellent at algebra and trust it!
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Find the point on the graph of the function π π₯ = (π₯ β 1)2 that is closest to the point (β5,3).
Minimize: π = (β5 β π₯)2+(3 β π₯ β 1 2)2
With πβ² =2π₯3β6π₯2+π₯+9
π₯4β4π₯3+π₯2+18π₯+29
With π₯ = β1, π¦ = (β1 β 1)2= 4.
The point (β1,4) on π π₯ = (π₯ β 1)2 is closest to (β5,3)
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On a given day, the flow rate πΉ (cars per hour)
on a congested roadway is πΉ =π£
22+0.02π£2 where
π£ is the speed of the traffic in miles per hour. What speed will maximize the flow rate on the road?
β’ βMaximize flow rateβ tells us to find the derivative.
πΉβ² =22+0.02π£2 1 βπ£(0.04π£)
(22+0.02π£2)2 =22β0.02π£2
(22+0.02π£2)2
πΉβ² = 0 when 22 β 0.02π£2 = 0 so π£ = 1100 β33 mph. Test points confirm this is a max.
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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 14 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area.
β’ Volume of a right circular cylinder and volume of a sphere combine to give 14.
β’ Minimize surface area.
β’ We need some formulas!
β’ π =4
3ππ3 + ππ2β = 14
β’ ππ΄ = 4ππ2 + 2ππβ 16
β’ π =4
3ππ3 + ππ2β = 14
β’ ππ΄ = 4ππ2 + 2ππβ
β’ To minimize surface area, we need to eliminate one of the variables. It seems easiest to eliminate the h. That is, we will solve the volume equation for h:
β =14 β
43
ππ3
ππ2=
14
ππ2β
4
3π
β’ Substitute into the surface area equation and then take the derivative.
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ππ΄ = 4ππ2 + 2ππβ
With β =14β
4
3ππ3
ππ2 =14
ππ2 β4
3π
Makes
ππ΄ = 4ππ2 + 2ππ14
ππ2β
4
3π
ππ΄ = 4ππ2 +28
πβ
8
3ππ2
ππ΄ =4
3ππ2 +
28
π
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ππ΄ =4
3ππ2 +
28
π
Taking the derivative:
ππ΄β² =8
3ππ β
28
π2
When the radius is 0 the surface area does not exist so letβs solve for ππ΄β = 0.
0 =8
3ππ β
28
π2
8
3ππ =
28
π2 and so 84 = 8ππ3 making π3 =84
8π or
π =84
8π
3β 1.495 cm.
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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. The total volume of the solid is 14 cubic centimeters. Find the radius of the cylinder that produces the minimum surface area.
β’ The radius that minimizes the surface area is
π =84
8π
3β 1.495 cm.
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A wooden beam has a rectangular cross section of height β and width π€. The strength π of the beam is directly proportional to the width and the square of the height by π = ππ€β2, where the k is the proportionality constant. What are the dimensions of the strongest beam that can be cut from a round log of diameter 20 inches?
β’ βStrongest beamβ tells us to maximize the strength equation given.
β’ The value k is a constant, but we have two other variables that we must get down to one.
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A wooden beam has a rectangular cross section of height β and width π€. What are the dimensions of the strongest beam, π = ππ€β2, that can be cut from a round log of diameter 20 inches?
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It seems as though we have a Pythagorean relationship with π€2 + β2 = 202.
Since our equation already has an β2, we can solve for it and substitute: β2 = 400 β π€2
So now,
π = ππ€(400 β π€2)
π = ππ€(400 β π€2)
Simplifying we have π = 400ππ€ β ππ€3
Taking the derivative will give πβ² = 400π β 3ππ€2
The critical numbers are when πβ = 0 so 0 = 400π β 3ππ€2
3ππ€2 = 400π
π€2 =400π
3π=
400
3
Therefore, π€ =400
3=
20
3=
20 3
3β 11.5 inches
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What are the dimensions of the strongest beam that can be cut from a round log of diameter 20 inches?
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With, π€ =400
3=
20
3=
20 3
3β 11.5 inches and
β2 = 400 β π€2, it must be that
β2 = 400 β400
3=
800
3
and so β =800
3β 16.3 inches.
The second derivative, π" = β6ππ€ which is negative for all positive widths; this is a max.
A man is in a boat 2 miles from the nearest point on the coast. He is to go to a point Q, located 3 miles down the coast and 1 mile inland. He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?
β’ We can break this into two parts:
β Time on water
β Time on land
β’ π· = ππ‘ so π‘ =π·
π
β’ Pythagoras 25
He can row at 2 miles per hour and walk at 4 miles per hour. Toward what point on the coast should he row in order to reach point Q in the least time?
β’ Water: π = π₯2 + 4 so Time = π₯2+4
2
β’ Land: π = (3 β π₯)2+1 = π₯2 β 6π₯ + 10 so
Time = π₯2β6π₯+10
4
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βleast timeβ means minimize the total time
Total Time = T =π₯2+4
2+
π₯2β6π₯+10
4
The minimum will occur at a critical number:
πβ² =1
2
1
2π₯2 + 4 β
12 2π₯
+1
4
1
2π₯2 β 6π₯ + 10 β
12 2π₯ β 6
πβ² =π₯
2 π₯2 + 4+
π₯ β 3
4 π₯2 β 6π₯ + 10
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Total Time = T =π₯2+4
2+
π₯2β6π₯+10
4
First derivative:
πβ² =π₯
2 π₯2 + 4+
π₯ β 3
4 π₯2 β 6π₯ + 10
Solving
0 =π₯
2 π₯2 + 4+
π₯ β 3
4 π₯2 β 6π₯ + 10
βπ₯
2 π₯2 + 4=
π₯ β 3
4 π₯2 β 6π₯ + 10
Square both sides to getβ¦
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Total Time = T =π₯2+4
2+
π₯2β6π₯+10
4
βπ₯
2 π₯2 + 4=
π₯ β 3
4 π₯2 β 6π₯ + 10
Square both sides to get π₯2
4(π₯2 + 4)=
π₯2 β 6π₯ + 9
16 π₯2 β 6π₯ + 10
Cross multiply to get 16π₯4 β 96π₯3 + 160π₯2
= 4π₯4 β 24π₯3 + 36π₯2 + 16π₯2 β 96π₯ + 144
Or π₯4 β 6π₯3 + 9π₯2 + 8π₯ β 12 = 0
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Total Time = T =π₯2+4
2+
π₯2β6π₯+10
4
Solving π₯4 β 6π₯3 + 9π₯2 + 8π₯ β 12 = 0
we get π₯ = 1 or π₯ β β1.11.
β’ Rowing a negative distance would be ridiculous so the man should row toward a point one mile down the coast to minimize his travel time.
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End of Lecture
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