1.The weight of a cylinder varies as the square of the diameter when the height is constant,it also varies as the height if the diameter remains constant.Two cylinders have their oheights is the ratio 49:72.Find the ratio of their diameter if the weight of the first is twice that of the second.2.Given that x varies as the cube of y when z is constant ,and x varies as z when y is constant and also that 5x=z when y=1,find the equation connecting x ,y an

1 圆柱重量问题高之比49:72 直径之比2:1 重量与体积成正比体积等于底面积×高 所以重量比例为49×4：72×1=49：18

1、当高度是常数，气缸重量随直径平方的变化；当直径保持常数时，它也随高度变化。如果第一个气缸的重量是第二个气缸的两倍时，两个气缸高度比试49:72时，求他们的直径比。
2、当Z是常数，且给出X随Y的立方变化（关系），且当Y是常数等于1,5X=Z，求X、Y、Z之间的等量关系
楼上答案正确，引用楼上：
1. Assume the ratio of diameter is x
(49:72) x^2=2:1
x^2= (2:1)/(49:72)=36:49
x=6:7
2 x= C y^3 z where C is a constant
at y=1, x= C z
since 5x=z, C=1/5
So, x =y^3 z/5

12:7 (注：下面的2指的是平方 第二个柱体的参数加了’以示区别)
质量=密度*体积=密度*πR2*h=密度*πD2*h/4
m/m'=(D2*h)/(D2'*h')=(49/72)*（D2/D2'）=2/1
∴D2/D2'=12:7
z=(5y)3=5x (3指的是立方)

1. Assume the ratio of diameter is x
(49:72) x^2=2:1
x^2= (2:1)/(49:72)=36:49
x=6:7
2 x= C y^3 z where C is a constant
at y=1, x= C z
since 5x=z, C=1/5
So, x =y^3 z/5

先翻译第一道题：当高不变时,圆柱体的重量与其直径的平方成正比；当直径不变时圆柱体的重量又与其高成正比.现有两个圆柱体,其高之比为49：72,且第一个圆柱体的重量为第二个的2倍.求这两个圆柱体的直径之比为多少?
依题意知圆柱体重量与直径、高的关系为M=k＊H＊D²（k为常数）,当H不变时,（kH）为常数,故符合题意；当D不变时,（kD²）为常数,故符合题意.所以,2=M1/M2=H1D1²/H2D2² 又H1/H2 =49/72 ,故D1²/D2² =144/49 ,得D1/D2 =12/7 .
翻译第二道题：当z不变时,x与y³成正比；当y不变时,x与z成正比.当y=1时,5x=z .求x,y,z之间的方程式.
依题意知x=k＊y³＊z ,(k为常数).当y=1时,5x=5 k＊1³＊z=z ,得k=1/5 .故所求方程式为：x=（1/5）＊y³＊z
希望能给你帮助!

V= PI * d^2/4 * h
V1=2*V2, H1:H2=49:72,所以 (D1:D2)^2 = (2*72):49
D1:D2=12:7
X＝k * y^3 * z
k=1/5
x= y^3 * z /5