1. Summer Assignment Answers

2. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing.

3. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing. We need to come up with two equations: One for the area. xy = 1000 One for the perimeter. 2y + x = L

4. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing. We need to come up with two equations: One for the area. xy = 1000 One for the perimeter. 2y + x = L Express L in terms of x alone. We need to solve for L without any y term, only x.

5. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing. We need to come up with two equations: One for the area. xy = 1000 y = 1000/x One for the perimeter. 2y + x = L Express L in terms of x alone. We need to solve for L without any y term, only x. 2(1000/x) + x = L

6. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing 2000/x + x = L Are there any restrictions on x?

7. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing 2000/x + x = L Are there any restrictions on x? 0 < x < size of the building.

8. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing 2000/x + x = L Make a graph of L versus x over a reasonable interval and find the value of x the results in the smallest value for L and find the value for L.

9. #9 A construction company wants to build a rectangular enclosure with an area of 1000 square feet by fencing in three sides and using its office building as the fourth side. Build an enclosure that uses the least amount of fencing 2000/x + x = L Make a graph of L versus x over a reasonable interval and find the value of x the results in the smallest value for L and find the value for L. Graph the equation above and find the minimum. Where the min occurs is your answer for x. The min value is your answer for L.

10. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2.

11. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. Make two equations, one for volume and one for cost.

12. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. Make two equations, one for volume and one for cost. 500 =  r 2 h C = (.02)2  r 2 + (.01)2  rh

13. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. Make two equations, one for volume and one for cost. 500 =  r 2 h C = (.02)2  r 2 + (.01)2  rh Express the cost C in terms of r.

14. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. Make two equations, one for volume and one for cost. 500 =  r 2 h C = (.02)2  r 2 + (.01)2  rh Express the cost C in terms of r. h = 500/(  r 2 ) C =.04  r 2 + 10/r

15. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. The top and bottom will be cut from squares. Make a new equation for C in terms of r.

16. #11 A soup company wants to manufacture a can in the shape of a right circular cylinder that will hold 500 cm 3 of liquid. The material for the top and bottom costs 0.02 cent/cm 2, and the material for the sides costs 0.01 cent/cm 2. The top and bottom will be cut from squares. Make a new equation for C in terms of r. 500 =  r 2 h C = (.02)8r 2 + (.01)2  rh Express the cost C in terms of r. h = 500/(  r 2 ) C =.16r 2 + 10/r

17. #9 As shown in the figure, a pendulum of constant length L makes an angle  with its vertical position. Express the height h as a function of the angle .

18. #9 As shown in the figure, a pendulum of constant length L makes an angle  with its vertical position. Express the height h as a function of the angle . When trying to get an angle involved, you need to use a trig function. To use a trig function, we need to have a right angle.

19. #9 As shown in the figure, a pendulum of constant length L makes an angle  with its vertical position. Express the height h as a function of the angle . When trying to get an angle involved, you need to use a trig function. To use a trig function, we need to have a right angle. cos  = (L – h)/L Lcos  = L – h h = L - Lcos 

20. #13 Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values.

21. #13a Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + 3x + 1

22. #13a Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + 3x + 1 If x > 0, the |x| = x f(x) = x + 3x + 1 x > 0

23. #13a Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + 3x + 1 If x > 0, the |x| = x f(x) = x + 3x + 1 x > 0 If x < 0, the |x| = -x f(x) = -x + 3x + 1 x < 0

24. #13a Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + 3x + 1 If x > 0, the |x| = x f(x) = x + 3x + 1 x > 0 If x < 0, the |x| = -x f(x) = -x + 3x + 1 x < 0 f(x) = 4x + 1 x > 0 f(x) = 2x + 1 x < 0

25. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases:

26. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases: x < 0 0 < x < 1 x > 1

27. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases: x < 0 |x| = -x |x – 1| = -( x – 1) = 1 – x 0 < x < 1 x > 1

28. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases: x < 0 |x| = -x |x – 1| = -( x – 1) = 1 – x 0 < x < 1 |x| = x |x – 1| = -(x – 1) = 1 – x x > 1

29. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases: x < 0 |x| = -x |x – 1| = -( x – 1) = 1 – x 0 < x < 1 |x| = x |x – 1| = -(x – 1) = 1 – x x > 1 |x| = x |x – 1| = x – 1

30. #13b Absolute Value and Piecewise functions. Express the function in piecewise form without using absolute values. f(x) = |x| + |x – 1| In this problem there are two points of interest; the zeros of each absolute value. We need to look at three cases: x < 0 |x| = -x |x – 1| = -( x – 1) = 1 – x 1 – 2x 0 < x < 1 |x| = x |x – 1| = -(x – 1) = 1 – x 1 x > 1 |x| = x |x – 1| = x – 1 2x – 1