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 was acceptable for posting, we need to determine the maximum value for the side length (x).
According to the given information, the sum of the length and girth of the parcel must not exceed 6 feet. The length of the parcel is x, and the girth is the perimeter of the cross-section, which is equal to 4 times the side length of the square (4x).
Therefore, the sum of the length and girth can be expressed as x + 4x = 5x feet.
Since the total must not exceed 6 feet, we can set up the equation:
5x ≤ 6
To find the maximum value for x, we divide both sides of the equation by 5:
x ≤ 6/5
Therefore, the maximum value for the side length (x) of the square cross-section is less than or equal to 6/5 feet.
To find the volume of the largest parcel, we use the formula for volume of a cube:
Volume = side length^3
Substituting the maximum value for the side length (x) into the formula:
Volume = (6/5)^3 = 216/125 cubic feet
So, the volume of the largest parcel with a square cross-section that was acceptable for posting is 216/125 cubic feet.
To find the volume of the largest square parcel that is acceptable for posting, we need to determine the dimensions that satisfy the given condition.
Let's assume the side length of the square cross-section is x feet.
According to the given condition, the sum of the length and girth (perimeter of the cross section) should not exceed 6 feet.
The length of the parcel is x feet, and the girth can be calculated by adding the four sides of the square. Since all sides of a square are equal, the girth would be 4 times the length of one side: 4x feet.
So, the total sum of the length and girth is x + 4x = 5x feet.
Since this sum should not exceed 6 feet, we can write the following inequality: 5x ≤ 6.
Now, let's solve this inequality to find the maximum value of x.
Dividing both sides of the inequality by 5: x ≤ 6/5.
So, the maximum value of x, and hence the side length of the square, is 6/5 feet.
To find the volume of the square parcel, we need to calculate the product of the side lengths: V = x^2.
Plugging in the maximum value of x, we have: V = (6/5)^2 = 36/25 square feet.
Therefore, the volume of the largest parcel with a square cross-section that is acceptable for posting is 36/25 square feet.