1. A 2 liter oil can is being designed in the shape of a right circular cylinder. What dimensions, in terms of centimeters, should the designer use in order to use the least amount of material? Hint: Recall 1 liter = 1000 ml = 1000 cm 3 When they say material what concept are they talking about: radius, area, volume? Volume of a cylinder = π r 2 h Surface Area of a cylinder = 2 π r 2 + 2π r h A. B. C. D. E. The designer should construct a can with the height of 2 centimeters and the radius of 318.47 centimeters in order to minimize cost. The designer should construct a can with the height of 8.6 centimeters and the radius of 8.6 centimeters in order to minimize cost. The designer should construct a can with the height of 20 centimeters and the radius of 10 centimeters in order to minimize cost. The designer should construct a can with the height of 243.74 centimeters and the radius of 6.83 centimeters in order to minimize cost. The designer should construct a can with the height of 13.66 centimeters and the radius of 6.83 centimeters in order to minimize cost.
2. A manufacturer wants to construct a box whose base length is 3 times the base width. The material to build the top and bottom cost 8 dollars per ft. 2 and sides 7 dollars per ft. 2. If the box must have a volume of 57 ft. 3, determine the dimensions that will minimize cost. Hint: Round to the hundreds place. You need to come up with a cost equation. Draw a picture! A. B. C. D. E. The manufacturer should construct a box with the length of 6.69 ft., the height of 0.51 ft., and the width of 2.23 ft. to minimize cost. The manufacturer should construct a box with the length of 33.25 ft., the height of 1.71 ft., and the width of 11.08 ft. to minimize cost. The manufacturer should construct a box with the length of 6.4 ft., the height of 4.18 ft., and the width of 2.13 ft. to minimize cost. The manufacturer should construct a box with the length of 6.69 ft., the height of 3.82 ft., and the width of 2.23 ft. to minimize cost. The manufacturer should construct a box with the length of 3 ft., the height of 3.82 ft., and the width of 2.23 ft. to minimize cost. 3. A woman is given a piece of cardboard that is 10 in. by 10 in. She wants to cut the corners of the cardboard so that she can fold the sides to make a box. Determine the height of the box that will give the maximum volume. A. The height of the box should be in. B. The height of the box should be in. C. The height of the box should be in. D. The height of the box should be in. E. The height of the box should be in.
Answers: 1. The designer wishes to minimize the amount of materials he will use. Write an equation representing the surface area of the can, including the top and bottom lids. Minimize Area (Principal equation): Write the secondary constant equation. In this case, the volume of the can remains constant throughout the problem. Remember that 1 L = 1000cm 3. Constant Volume ---------------------------------------------------------------------------------------------------------- In order the get the surface area formula in terms of a single variable, r, solve the constant volume equation for h and make a substitution into the surface area equation. Take the derivative of the surface area equation. Set equal to 0 and solve for r to obtain the critical points.
At the critical number 0, r is undefined and cannot be used. Check the critical number 6.83 to confirm that it is a minimum value. Choose two numbers. One number must be greater than the critical number r, and one number must be less than. ------------------6.83 ++++++++++++ 6.83 is a minimum because of a sign change from negative to positive with the tested points. The radius of the the oil can should be 6.83 cm in order to minimize cost. The height must now be found. Substitute the measure of the radius into the equation defined as h and simplify. The designer should construct a can with the height of 13.66 centimeters and the radius of 6.83 centimeters in order to minimize cost.
2. Write an equation representing the surface area of the box. Also define variables. In this case, H represents the box's height, W represents the box's width, and L represents the box's length. Merge the facts that L = 3W with the way to find the surface area: The manufacturer wishes to minimize the cost of the materials that he will use. Write an equation representing the surface area of the box, including the top and bottom lids. Minimize Cost (This is the principal equation): Since the top and bottom cost 8 dollars per ft. 2 and sides 7 dollars per ft. 2 the Surface area formula becomes a cost formula of: Write the constant equation. In this case, the volume of the box remains constant throughout the problem. Constant Volume (This is the secondary equation): In order the get the cost formula in terms of a single variable, W, solve the constant volume equation for H and make a substitution into the cost equation. Take the derivative of the cost equation.
Set equal to 0 and solve for W to obtain the critical points. Since the derivative is undefined at W = 0, 0 is also a critical number. However, at 0, C is undefined in the original function; therefore, 0 cannot be used. Check the critical number 2.23 to confirm that it is a minimum value. Choose two numbers. One number must be greater than the critical number W, and one number must be less than. 2.23 is a minimum value because there is a sign change from negative to positive with the tested points. The width of the box should be 2.23 ft. in order to minimize cost. The height must now be found. Substitute the measure of the width into the equation defined as H and simplify. The length of the box must also be found. Remember that L = 3W. Substitute the measure of the width into the equation defined as L and simplify. The manufacturer should construct a box with the length of 6.69 ft., the height of 3.82 ft., and the width of 2.23 ft. to minimize cost.
3. Draw a digram. Write an equation representing the volume of the box. Also define any variables to be used. In this case, H represents the box's height. Maximize Volume Take the derivative of the volume equation. Set equal to 0 and solve for H to obtain the critical point(s).
Check the critical numbers to determine which number is the maximum value. Choose two numbers. One number must be greater than the critical number H, and one number must be less than. is the maximum value because there is a sign change from positive to negative with the tested points. The height of the box should be in.