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.)

Question

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.)

Answer 1

x^2 y = V

G = 4x+y = 6
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y = 6-4x

x^2 (6-4x) = V

now do the complete the square thing like I did in your other problem to find the vertex of the parabola

Answer 2

The perimeter of a square is given by 4 times the length of one side, which means the girth of the parcel, in this case, is 4x.

According to the given condition, the sum of the length and girth must not exceed 6 feet.

So, the length + girth ≤ 6

Substituting the values: x + 4x ≤ 6

Combining like terms: 5x ≤ 6

Dividing both sides by 5: x ≤ 6/5

Therefore, the side of the square cross-section must be less than or equal to 6/5 feet.

To find the maximum volume, we need to find the maximum value of x.

Since the cross-section is a square, the volume of the parcel is given by V = x^2.

Let's find the value of x that maximizes V. Taking into account that x ≤ 6/5, we should use the maximum value:

x = 6/5

Plug this value of x into the volume formula:

V = (6/5)^2

V = 36/25

Therefore, the largest parcel with a square cross-section acceptable for posting has a volume of 36/25 cubic feet.

Answer 3

To find the volume of the largest parcel of square cross-section acceptable for posting, we need to determine the value of x that maximizes the volume within the given constraints.

Given that the sum of the length and girth must not exceed 6 feet, we can write the equation as:

x + 4x ≤ 6

Simplifying the equation, we have:

5x ≤ 6

Dividing both sides by 5:

x ≤ 6/5

Now, we can express the volume of the parcel as:

Volume = x² * x = x³

To maximize the volume, we need to find the maximum value of x within the given constraints. Since x must also be positive, we can take the maximum value of x to be the value obtained from the equation:

x = 6/5

Substituting this value into the volume equation, we have:

Volume = (6/5)³ = 216/125 cubic feet

Therefore, the volume of the largest parcel of square cross-section acceptable for posting is 216/125 cubic feet.