el e R R e Kt S 45. MAXIMUM VOLUME A sector with central angle 6 is cut from a circle of radius 12 inches (see figure), and the edges of the sector are brought together to form a cone. Find the magnitude of 6 such that the volume of the cone is a 1 maximum. V = g}trzh Volume of Cone = l xr'h 3 Hint: While this problem can feel daunting, the key to solving it is to name things differently. In class, when dealing with the volume of a cube, we said V = [ - W - H. But I assumed the height to be unique and the length and width to be unknowns by naming them both x. Hence, V=1- W -H = x2H . 1 did this to relate the Volume to the Area, which is all we were given. Here, relate the height of the cone by means of the radii (plural of radius) of the cone, r,with that of the circle, 7, via Pythagoras. Once you get through the first derivative and find the critical values. Recall before doing this, you need a suitable domain. Also, as a hint: (96 r2) can be broken into a difference of squares. If you cannot see it, take the root of 96. The total area of a cone is given by 77> + nrhy, where ar? is the area of the circular base/ opening, zrh; lateral surface area, and i, = cone's lateral height. The area of the sector of a circle is given by @7r%. You now have all the formulas, and hints, to complete this problem. Here, what is hardest are the applications, which requires us to remember precalculus, geometry, and a lot of algebra. has a rectangular cross section of height h and width w (see figure). The strength S of the beam is directly proportional to the width and the square of the height. What are the dimensions of the strongest beam that can be cut from a round log of diameter 20 inches? (Hint: S = kh*w, where k is the proportionality constant.) [Double hint: isolate the right variable to replace on the primary equation. Also, since testing extreme values here would not be helpful, use the second derivative to determine where you have a max or a min