✔ 最佳答案
(1)
(x^2)(h) = 20 => h = 20/x^2
設 $C 為成本,則 C = 2(500)(x^2) + 4(800)(xh)
C = 1000x^2 + 3200xh = 1000x^2 + 3200x(20/x^2) = 1000x^2 + 64000/x
dC/dx = 2000x - 64000/x^2 => d^2C/dx^2 = 2000 + 128000/x^3
解 dC/dx = 0,可得 x = 32^(1/3) = 2[4 ^ (1/3)]
把 x = 2[4 ^ (1/3)] 代入 d^2C/dx^2 可得 d^2C/dx^2 > 0
所以最小成本 = 1000[32^(1/3)]^2 + 64000 / 32^(1/3) = 24000[2^(1/3)]
(2)
∠POC = 2Ф
L = AP + PC = sin2Ф + sin2Фcosec2Ф = 2cosФ + sin2Ф
dL/dФ = -2sinФ + 2cos2Ф => d^2L/dФ^2 = -2cosФ - 4sin2Ф
解 dL/dФ = 0, 可得 2(sinФ)^2 + sinФ - 1 = 0
sinФ = 1/2 或 -1(捨去) => Ф = pi/6
把 Ф = pi/6 代入 d^2L/dФ^2 可得 d^2L/dФ^2 > 0
因為 0 < Ф < pi/2 中只有一個轉向點,
所以 L 的極大值 = 2cos(pi/6) + sin2(pi/6) = 3^(1/2) + 3^(1/2)/2 = 3[3^(1/2)]/2