Multiple choice

The sum of the legs of a right triangle is 9 cm. When the triangle rotates about one of the legs, a cone results which has the maximum volume. Find the lateral area of the cone.

  1. $18 \sqrt{5} \, cm^2.$
  2. $5 \sqrt{18}\pi \, cm^2.$
  3. $18 \sqrt{5}\pi \, cm^2.$
  4. $5 \sqrt{18} \, cm^2.$
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C Correct answer
Explanation

Let legs be x and 9-x. Volume V = (1/3) * pi * r^2 * h. If rotated about leg h, r = 9-h. V = (1/3) * pi * (9-h)^2 * h. dV/dh = (pi/3) * ((9-h)^2 - 2h(9-h)) = 0. (9-h)(9-h-2h) = 0. h = 3, r = 6. Lateral area = pi * r * l = pi * r * sqrt(r^2 + h^2) = pi * 6 * sqrt(36 + 9) = pi * 6 * sqrt(45) = pi * 6 * 3 * sqrt(5) = 18 * sqrt(5) * pi.

AI explanation

For a right triangle with legs r and h rotated about h to form a cone, the sum of the legs is r + h = 9, so h = 9 - r. The volume of a cone is V = (1/3)πr^2h, so substituting h gives V = (1/3)πr^2(9 - r) = (1/3)π(9r^2 - r^3). To maximize this volume, set the derivative of (9r^2 - r^3) to zero, yielding 18r - 3r^2 = 0, which gives r = 6 cm and h = 3 cm. The slant height is l = sqrt(6^2 + 3^2) = sqrt(45) = 3sqrt(5) cm. The lateral surface area is πrl = π * 6 * 3sqrt(5) = 18sqrt(5)π cm^2.

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