the maximum volume will be 128 in. 3 TRY YOUR TURN 3 EXAMPLE 4 Minimizing Area A coffee company wants to manufacture cylindrical aluminum coffee cans with a volume of 1000 cm (1 liter). What should the radius and height of the can be to minimize the amount of aluminum used? APPLY IT SOLUTION The two variables in this problem are the radius and the height of the can, which we shall label r and h, as in Figure 9. Minimizing the amount of aluminum used requires minimizing the surface area of the can, which we will designate S. The surface area consists of a top and a bottom, each of which is a circle with an area rr, plus the side. If the side were sliced vertically and unrolled, it would form a rectangle with height h and width equal to the circumference of the can, which is 2r. Thus the surface area is given by S = 2™r + 2mrh. The right side of the equation involves two variables. We need to get a function of a single variable. We can do this by using the information about the volume of the can: vie nd V = ar?h = 1000. FIGURE 9 (Here we have used the formula for the volume of a cylinder.) Solve this for h: 1000 h = %3D TTr (Solving for r would have involved a square root and a more complicated function.) We now substitute this expression for h into the equation for S to get 0.ovigonod asa ribi 1000 S = 2Tr2 + 2r 2000 2mr + There are no restrictions on r other than that it be a positive number, so the domain of S is (0, 0). Find the critical points for S by finding dS/dr, then solving the equation dS/dr = 0 for r. dS 2000 dr Add 2000 to both sides and multiply by r². 4пr3 2000 500 p3 = TT Take the cube root of both sides to get d) 500\/3 - 5.419 TT 14.2 Applications of Extrema 767 14. centimeters. Substitute this expression into the equation for h to get 1000 - 10.84 %3D T(5.419)2 centimeters. Notice that the height of the can is twice its radius. There are several ways to carry out Step 6 to verify that we have found the minimum. Because there is only one critical number, the critical point theorem applies. Method I Critical Point Theorem Verify that when r< 5.419, then dS/dr < 0, and when r> 5.419, then dS|dr > 0. Since the function is decreasing before 5.419 and increasing after 5.419, there must be a relative minimum at r = 5.419 cm. By the critical point theorem, there is an absolute minimum there. with First Derivative Test %3D Method 2 We could also use the critical point theorem with the second derivative test. Critical Point Theorem d?s 4000 with Second Derivative Test 4т + 3 %3D 2 dr3 Notice that for positive r, the second derivative is always positive, so there is a relative minimum at r = 5.419 cm. By the critical point theorem, there is an absolute minimum there. Method 3 We could also find the limit as the endpoints are approached. Limits at Endpoints limS = = ∞ lim S = O0 %3D r 00 The surface area becomes arbitrarily large as r approaches the endpoints of the domain, so the absolute minimum surface area must be at the critical point. YOUR TURN 4 Repeat Exam- ple 4 if the volume is to be 500 cm?. The graphing calculator screen in Figure 10 confirms that there is an absolute minimum at r = 5.419 cm. TRY YOUR TURN 4 %3D Notice that if the previous example had asked for the height and 2000