✔ 最佳答案
1.
設此兩正數為 x , 60-x
乘積 P(x) = x(60-x) = - x^2 + 60x
P ' = - 2x + 60 = 0 , 故 x = 30 , 在此點有極值.
P '' = - 2 < 0 , 故該極值為極大值.
Ans : 30 , 30
2.
令 A = ( 1 , 2 )
且令 B = ( p , p^2/4 ) 在此曲線上, 且最接近A .
y ' = 2x/4 = x/2
y ' ( p ) = p/2 , 即B點的切線斜率.
設 AB 之斜率為 m
因為B最接近A, 所以 AB 垂直於 B點的切線, 故得:
m * (p/2) = - 1
m = - 2/p
由A,B座標 與 點斜式公式得:
m = ( p^2/4 - 2 ) / ( p - 1 ) = - 2/p
- 2p + 2 = p^3/4 - 2p
p^3 = 8
p = 2
B = ( p , p^2/4 ) = ( 2 , 1 )
Ans: ( 2 , 1 )
3.
設長寬高分別為 L , W , H
L = 3W
L * W * H = 486
3W^2 * H = 486
H = 486 / (3W^2) = 162 / W^2
A
= 2LH + 2WH + LW
= 6WH + 2WH + 3W^2
= 8W( 162 / W^2) + 3W^2
= 1296/W + 3W^2
≡ A(W)
A ' = - 1296/W^2 + 6W = 0 , 解得 W = 6 , 在此點有極值
A '' = 2592/W3 + 6
A '' (6) = 2592/6^3 + 6 = 2592/216 + 6 = 12 + 6 = 18 > 0 , 故此極值為極小值
A(6) = 1296/W + 3W^2 = 1296/6 + 3*6^2 = 216 + 108 = 324
Ans: 324 in^2
4.
設圓錐高度為 H , 底面半徑為 R ; 其內接圓柱體的高度為 h , 底面半徑為 r .
此圓錐為直角圓錐, 故由側視圖看, 頂角為90° , 故兩側底角為45° , 如底下附圖所示.
故可推得 :
H = R 且 R - r = h
內接圓柱體的體積 V
= π r^2 h
= π r^2 ( R - r )
= πRr^2 - πr^3
V ' = 2πRr - 3πr^2 = π r ( 2R - 3r ) = 0 , 解得 r = 0 (不合) , 2R/3
V '' = 2πR - 6πr
V '' ( 2R/3 ) = 2πR - 6π(2R/3) = 2πR - 4πR < 0 , 故該點為極大值
當 r = 2R/3 , 圓柱體體積最大 ,
此時 h = R - r = R - 2R/3 = R/3 = H/3
Ans: 圓柱體高度 = (1/3)*圓錐高度 = (1/3)*圓錐底面半徑