中五變分應用題 一高模型的高及重量分別是x cm 及 y g。已知y是兩部分的和，一部分隨x正變，另一部分隨x^2正變。當x = 11時，y = 880 ; 且當x = 12時，y = 1044。?

y = px + qx^2 , for some p > 0 , q > 0
y(11) = 11p + 121q = 880 ..... (1)
y(12) = 12p + 144q = 1044 .....(2)
(2) - (1)得:
p = 164 - 23q , 代回(1)得:
11( 164 - 23q ) + 121q = 880
q = 7
p = 164 - 23*7 = 3
故 y = 3x + 7x^2

(a)
y(13) = 3*13 + 7*13^2 = 1222

(b)
2584 = 3x + 7x^2
7x^2 + 3x - 2584 = 0
x
= { - 3 ± √ [ 3^2 - 4*7*(-2584) ] } / (2*7)
= ( - 3 ± 269 ) / 14 , 負不合
= 19

(c)
體積是(a)模型的8倍
8^(1/3) = 2
故高度是(a)模型的2倍 , 2*13 = 26 , 即高度為 26 cm
y(26) = 3*26 + 7*26^2 = 4810

Ans:
(a) 1222 g
(b) 19 cm
(c) 4810 g