The size of a parcel despatched through the post used to be limited by the fact that the sum of its length and girth (perimeter of the cross section) must not exceed 6 feet. What was the volume of the largest parcel of square cross- section which was acceptable for posting? (Let the cross-section be a square of side x feet.)
To find the volume of the largest parcel of square cross-section that is acceptable for posting, we need to first understand the dimensions and constraints given.
The constraint states that the sum of the length and girth (perimeter of the cross-section) of the parcel must not exceed 6 feet.
Let's break this down:
1. The length of the parcel is the same as the side length of the square cross-section, which is given as x feet.
2. The girth refers to the perimeter of the cross-section, which for a square is equal to 4 times the side length, or 4x feet.
According to the constraint, the sum of the length and girth should not exceed 6 feet:
x + 4x ≤ 6
Now we can solve for x:
5x ≤ 6
Divide both sides by 5:
x ≤ 6/5 or 1.2 feet
Since we are looking for the largest parcel, we want to maximize the volume. The volume of a square cross-section parcel is simply the area of the square multiplied by its length (side length), so:
Volume = x^2 * x = x^3
To find the largest possible volume, we substitute the value of x that we obtained from the constraint:
Volume = (1.2)^3 = 1.728 cubic feet
Therefore, the volume of the largest parcel of square cross-section that is acceptable for posting is 1.728 cubic feet.
v = x^2y
4x+y <= 6
see where that takes you.