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?
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.
To find the maximum length of the ribbon required, we need to calculate the perimeter of the rectangular present.
Let's assume the length, width, and height of the rectangular present as l, w, and h respectively.
The volume of the present is given as 2592cm^3. Since it is a rectangular box, the volume can be calculated as:
Volume = length * width * height
2592 = l * w * h
We are given that the ribbon is tied lengthwise and then crosswise around the present. This means that the length of the ribbon required will be equal to the perimeter of the rectangular box plus an extra length for the bow.
The perimeter of a rectangle can be calculated as:
Perimeter = 2 * (length + width)
Given that there is an extra 30cm length for the bow, the maximum length of the ribbon required can be expressed as:
Maximum Length of Ribbon = Perimeter + 30
Now, let's substitute the formulas for Perimeter and the given volume:
Maximum Length of Ribbon = 2 * (length + width) + 30
We can rewrite this equation using the volume formula as:
Maximum Length of Ribbon = 2 * (l + w) + 30
Now we have an equation with three variables (l, w, and h) and one equation (2592 = l * w * h).
To find the maximum length of the ribbon required, we need to solve these equations simultaneously. However, we need additional information or constraints on the dimensions of the rectangular present to provide a specific answer.