A jewelry designer plans to incorporate a component made of gold in the shape of a frustum of a cone of height 1 cm and fixed lower radius r (Figure 43). The upper radius x can take on any value between 0 and r. Note that x = 0 and x = r correspond to a cone and cylinder, respectively. As a function of x, the surface area (not including the top and bottom) is S (x) = πs(r + x), where s is the slant height as indicated in the figure. Which value of x yields the least expensive design [the minimum value of S (x) for 0 ≤ x ≤ r]?

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(a) Show that S (x) = n(r + x) √√1 + (r − x)². (b) Show that if r < √2, then S is an increasing function. Conclude that the cone (x = 0) has minimal area in this case. (c) Assume that r> √2. Show that S has two critical points x₁ < x2 in (0,r), and that S (x₁) is a local maximum, and S (x₂) is a local minimum. (d) Conclude that the minimum occurs at x = 0 or x₂. (e) Find the minimum in the cases r = 1.5 and r = 2. (1) Challenge: Let c = √(5 + 3√3)/4 ≈ 1.597. Prove that the minimum occurs at x = 0 (cone) if √2 c.