Asked by Julian
A ribbon is to be tied lengthwise and then crosswise around a square based rectangular present leaving an extra 30cm length for a bow. If the volume of the present is 2592cm^3, what is the maximum length of the ribbon required?
Answers
Answered by
Steve
the dimensions of the box are x,x,y where x^2 y = 2592, so y = 2592/x^2
Around the square, the distance is 4x
Around the length, the distance is 2x+2y
So, we want the minimum length. No maximum is possible, as the package may be arbitrarily long and thin.
So, the length z is
z = 4x+2x+2y+30 = 6x+5184/x^2 + 30
dz/dx = 6 - 10368/x^3
dz/dx=0 when x = 12
That makes z(12) = 138cm as the shortest ribbon which will do the job.
Around the square, the distance is 4x
Around the length, the distance is 2x+2y
So, we want the minimum length. No maximum is possible, as the package may be arbitrarily long and thin.
So, the length z is
z = 4x+2x+2y+30 = 6x+5184/x^2 + 30
dz/dx = 6 - 10368/x^3
dz/dx=0 when x = 12
That makes z(12) = 138cm as the shortest ribbon which will do the job.
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