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13. The volume of a rectangular box is given by the function \( V(w) = (60 - 4w)w^2 \). What is a reasonable domain for the function in this situation? Express the domain as an inequality, in interval notation, and in set notation.

14. Sketch a graph of the function in Item 13 over the domain that you found. Include the scale on each axis.

15. Use a graphing calculator to find the coordinates of the maximum point of the function given in Item 13.

16. What is the width of the box, in inches, that produces the maximum volume?

17. Reason abstractly. An architect uses a cylindrical tube to ship blueprints to a client. The height of the tube plus twice its radius must be less than 60 cm.

a. Write an expression for \( h \), the height of the tube, in terms of \( r \), the radius of the tube.

b. Write an expression for \( V \), the volume of the tube, in terms of \( r \), the radius of the tube.

c. Find the radius that produces the maximum volume.

d. Find the maximum volume of the tube.

Answer :

13. The reasonable domain for the function V(w) = (60-4w)w^2 is when the volume is greater than 0.

14. The graph will look like a parabola with a maximum point at w = 7.5.

15.The coordinates of the maximum point are (7.5, 506.25).
16. The width of the box that produces the maximum volume is 7.5 inches, as found in the previous question.

17. a. The expression for the height of the tube in terms of the radius is h = 60 - 2r.

This means that the values of w must be between 0 and 15, since the volume becomes negative when w is greater than 15. Therefore, the domain can be expressed as an inequality as 0 < w < 15, in interval notation as (0, 15), and in set notation as {w | 0 < w < 15}.

14. To sketch a graph of the function V(w) = (60-4w)w^2 over the domain (0, 15), we can plot points at different values of w and connect them with a smooth curve. The scale on each axis can be 1 unit per grid line.

15. Using a graphing calculator, we can find the coordinates of the maximum point of the function V(w) = (60-4w)w^2 by using the maximum function.

b. The expression for the volume of the tube in terms of the radius is V = πr^2h = πr^2(60 - 2r).


c. To find the radius that produces the maximum volume, we can take the derivative of the volume function and set it equal to 0. This gives us 2πr(60 - 4r) = 0. Solving for r, we get r = 7.5 cm.


d. The maximum volume of the tube is V = π(7.5)^2(60 - 2(7.5)) = 1265.49 cm^3.

To know more about interval notation click on below link:

https://brainly.com/question/29531272#

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