Problem of the Day calculator If fx ln

Problem of the Day calculator! If f(x) = { ln x 2 x ln 2 No for 0 < x < 2 for 2 < x < 4 then lim f(x) is x⇒ 2 A) ln 2 C) ln 16 nonexistent B) ln 8 D) 4 E)

Problem of the Day If f(x) = { ln x 2 x ln 2 for 0 < x < 2 for 2 < x < 4 then lim f(x) is x⇒ 2 A) ln 2 C) ln 16 nonexistent B) ln 8 D) 4 E) limit from left does not equal limit from right at 2

Optimization Problems One of the most common uses of calculus is to determine minimum and maximum values. Some greatest profit examples are least cost least time greatest voltage optimum size greatest strength greatest distance greatest volume least material least size

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 1 - Guess and check x h x surface area = area of base + 4 (area of a side) 108 = (x)(x) + 4(x)(h) 108 = x + 4 xh 2

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? x x Method 1 - Guess and h check 108 = x + 4 xh 2 volume = x h 2

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 1 - Guess and h 108 = x + 4 xh check 2 x volume = x h x length of sideheightvolume 3 4 5 6 7 8 2

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 1 - Guess and h 108 = x + 4 xh check 2 x x length of side 3 4 5 6 7 8 volume = x h 2 volume 8. 25 74. 25 5. 75 92 4. 15 103 ¾ 108 3 2. 11 103¼ 1. 375 88 height

Any problems with this method? Method 1 - Guess and h 108 = x + 4 xh check 2 x x length of side 3 4 5 6 7 8 volume = x h 2 volume 8. 25 74. 25 5. 75 92 4. 15 103 ¾ 108 3 2. 11 103¼ 1. 375 88 height

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 2 - Use Calculus x h surface area = area of base + 4 (area of a x side) 108 = x 2 + 4 xh volume = x 2 h What is to be maximized? What is this in Calculus?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 2 - Use Calculus x surface area = area of base + 4 (area of a h 2 108 = x + 4 xh side) x 2 volume = x h What is to be maximized? What is this in Calculus? Number Volume Critical And Critical Numbers come from?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 2 - Use Calculus x surface area = area of base + 4 (area of a h 2 108 = x + 4 xh side) x 2 volume = x h What is to be maximized? Volume Critical What is this in Calculus? Number And Critical Numbers come from? First Derivative Set Equal to Zero

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 2 - Use Calculus x surface area = area of base + 4 (area of a h 2 108 = x + 4 xh side) x 2 volume = x h How can we eliminate a variable?

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume? Method 2 - Use Calculus x surface area = area of base + 4 (area of a h 2 108 = x + 4 xh side) x 2 volume = x h How can we eliminate a variable? Use the equation not being maximized to solve for one of the variables and substitute the result into the equation being maximized.

Procedure to Maximizing or Minimizing 1) Develop 2 equations 2) Use equation not being maximized to eliminate a variable 3) Determine feasible domain for x 4) Differentiate 5) Find critical numbers 6) Evaluate at endpoints of domain and at critical numbers 7) Pick value that gives the largest or smallest value

1) Develop 2 108 = x equations + 4 xh 2 volume = x h 2 2) Use equation not being maximized to 108 = x +eliminate 4 xh a variable 2 h = 108 - x 4 x 2 V=x h 2 2 V= (x ) 108 - x 4 x 3 V = 27 x - ¼x 2

3) Determine feasible domain for x 1) x must be greater than zero 2) The largest x would be when the height is zero. 2 2 Then 108 = x + 0 and x = √ 108 Thus 0 < x < √ 108 4) Differentiate V = 27 x - ¼x d 2 V = 27 - ¾x d x 3

5) Find critical numbers d. V = 27 - ¾x d x 2 0 = 27 - ¾x 2 3 2 x = 108 x = ± 6 6) Evaluate at endpoints of domain and f(0) = at 0 critical numbers f(6) = 108 f(√ 108) = 0

7) Pick value that gives the largest or value f(0) =smallest 0 f(6) = 108 maximum f(√ 108) = 0 so x = 6 and h = 3 dimensions are 6 x 3

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum.

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum. 1) Develop 2 equations 2) Use equation not being maximized to eliminate a variable

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum. 3) Determine feasible domain for x 4) Differentiate 5) Find critical numbers

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum. 6) Evaluate at endpoints of domain and at critical numbers 7) Pick value that gives the largest or smallest value

Find 2 positive integers whose product is 192 and the sum of the first plus three times the second is a minimum. The numbers are 24 and 8

Find the length and width of a rectangle that has an area of 64 and a minimum perimeter.

Find the length and width of a rectangle that has an area of 64 and a minimum perimeter. Area = l w. 64 = l w 64/w = l. Perimeter = 2 l + 2 w Perimeter = 2(64/w) + 2 w y' = -128/w + 2 2 0 = -128/w + 2 8=w 2 Length = 8 and width = 8

Find the point on the graph of the function f(x) = 2 x that is closest to the point (2, ½).

Find the point on the graph of the function f(x) = 2 x that is closest to the point (2, ½). y = x 2 d 2 2 istance = √(2 - x) + (½ - y) distance = (x 4 - 4 x + 4. 25)½ distance ' = ½(x 4 - 4 x + 4. 25)-½ (4 x 3 - 4) 0 = ½(x - 4 x + 4. 25) (4 x - 4) 1=x 4 -½ 3

A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank?

A tank with a rectangular base and rectangular sides is to be open at the top. It is to be constructed so that its width is 4 meters and its volume is 36 cubic meters. If building the tank costs $10 per square meter for the base and $5 per square meter for the sides, what is the cost of the least expensive tank? $330

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