Let alpha the angle that: sin(alpha) = 2/y and cos(alpha) = 1/x, then x+y = 1/cos(alpha) + 2/sin(alpha). We optimize (minimize) the last function of alpha by derive respect alpha, and then f '(alpha)=0 to obtain: tan^3(alpha)=2. Now: x + y = (2*cos(alpha)+sin(alpha))/(sin(alpha)*cos(alpha)) = (2+tan(alpha)/sin(alpha). sin^2(alpha)=(2^(2/3))/(1+2^(2/3)), then:
Answer is B.
ReplyDeleteLet alpha the angle that:
sin(alpha) = 2/y and cos(alpha) = 1/x, then x+y = 1/cos(alpha) + 2/sin(alpha).
We optimize (minimize) the last function of alpha by derive respect alpha, and then f '(alpha)=0 to obtain: tan^3(alpha)=2.
Now: x + y = (2*cos(alpha)+sin(alpha))/(sin(alpha)*cos(alpha)) = (2+tan(alpha)/sin(alpha).
sin^2(alpha)=(2^(2/3))/(1+2^(2/3)), then:
x+y= (1+2^(2/3))^(3/2).
el-perfecto !
ReplyDeleteGeneral solution:
ReplyDeleteIf a is the distance between first and second paralleles and b is the distance between second a third, then:
min(x+y) = [a^(2/3)+b^(2/3)]^(3/2)
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