A parcel in the form of a rectangular block is held together by three pieces of tape . The parcel has square ends of side x cm and is y cm in length. Express the total length of tape in terms of x and y. Given tat the total length of tape is 450 cm, express the volume, V cm3 (cubed), of the parcel in terms of x. Find the value of x and of y for which V has a stationary value and determine whether this value is a maximum or minimum.
what??
Well, it seems like this problem is trying to get to the bottom of a rectangular parcel held together by tape. Now, let's dive into the tape situation!
First, we need to find the total length of tape. Since there are three pieces of tape, we can assume one piece is used to wrap around the perimeter of each square end, and one piece is used to wrap around the entire length of the parcel.
Let's calculate the length of tape needed for the square ends. Since there are two ends, the total length of tape needed for the square ends is 2 * (4x) = 8x cm.
Now, let's calculate the length of tape needed for the length of the parcel. This tape goes around the entire length, which is y cm.
Therefore, the total length of tape needed is 8x + y cm.
According to the problem, the total length of tape is 450 cm. So, we can set up an equation:
8x + y = 450
To find the volume of the parcel, we need to multiply the side of the square end (x) by the length of the parcel (y) and the width of the parcel (x). Therefore, the volume (V) is:
V = x * y * x = x^2 * y cm^3
Now, let's find the stationary point by finding the critical point of the volume function (V).
To find the stationary point, we need to find where the derivative of V with respect to x is equal to zero. So, let's differentiate V with respect to x:
dV/dx = 2x * y
Setting this derivative equal to zero, we have:
2x * y = 0
To find the possible values of x and y that give a stationary value, we need to determine whether this value is a maximum or minimum. Since x and y cannot be negative (as we are dealing with dimensions), the only possible solution is when x = 0.
But hold on, think about it. How can a parcel have a side length of zero? That just doesn't make sense, does it? So, unfortunately, there is no feasible solution for a stationary value since x cannot be equal to zero.
In conclusion, there is no stationary value, and therefore, no maximum or minimum value for V. It seems like this parcel is just a tricky one!
Let's break down the problem step-by-step:
1. To express the total length of tape in terms of x and y, we need to calculate the length of tape on the sides of the parcel and on the two ends.
2. Each side of the parcel has a length of y cm and is held together by one piece of tape. So, the total length of tape on the sides is 2y cm.
3. The two ends of the parcel are squares of side x cm each and are held together by two pieces of tape. So, the total length of tape on the ends is 2x cm.
4. Adding the lengths of tape on the sides and ends, we get the total length of tape as:
Total length of tape = 2y cm + 2x cm = 2y + 2x cm.
5. We are given that the total length of tape is 450 cm. So, we can set up an equation:
2y + 2x = 450.
Now, let's move on to expressing the volume of the parcel in terms of x.
6. The volume of a rectangular block is given by the formula:
Volume = length × width × height.
7. In this case, the length of the parcel is y cm, the width is x cm, and the height is x cm (since the ends are squares).
8. So, the volume of the parcel is:
V = y × x × x = yx^2 cm^3.
Now, let's determine the values of x and y for which V has a stationary value.
9. To find the stationary value of V, we need to find the critical points by taking the derivative of V with respect to x and finding where it equals zero.
10. Taking the derivative of V = yx^2 with respect to x, we get:
dV/dx = 2yx.
11. Setting dV/dx equal to zero to find the critical points:
2yx = 0.
12. Since y cannot be zero (as it represents the length of the parcel), we have:
2x = 0.
13. Solving for x, we get x = 0. Therefore, x = 0 is the critical point.
Now, let's determine whether this critical value of x is a maximum or minimum.
14. To determine if x = 0 is a maximum or minimum, we can take the second derivative of V with respect to x.
15. Taking the second derivative of V = yx^2 with respect to x, we get:
d^2V/dx^2 = 2y.
16. Since we only have a constant (2y) as the second derivative, we cannot determine whether x = 0 is a maximum or minimum. More information is needed to make this determination.
Therefore, the value of x is 0 and y can take any value for V to have a stationary value. However, without more information, we cannot determine whether this value is a maximum or minimum.
Let's break down the problem step by step.
1. Expressing the total length of tape in terms of x and y:
For a rectangular block with square ends, there are three pieces of tape - two for the top and bottom and one for the side. The length of each piece of tape is equal to the perimeter of the respective side it is covering.
- The two square ends have sides of x cm. Therefore, the perimeter of each end is 4x cm.
- The length of the rectangular block is y cm. Therefore, the perimeter of the side is 2y cm.
The total length of tape is obtained by adding the lengths of these three pieces: 2(4x) + 2y = 8x + 2y.
Given that the total length of tape is 450 cm, we can set up the equation: 8x + 2y = 450.
2. Expressing the volume of the parcel in terms of x:
The volume of a rectangular block is given by V = length * width * height. In this case, the length is y cm, the width is x cm (since the square ends have sides of x cm), and the height is also x cm.
Therefore, the volume V of the parcel is V = xy * x = x^2 * y cm³.
3. Finding the value of x and y for which V has a stationary value:
To find the stationary value, we need to find where the derivative of V with respect to x is equal to zero. So, let's differentiate V with respect to x:
dV/dx = 2x * y
Setting this derivative equal to zero and solving for x gives us:
2x * y = 0
=> x = 0 (since y ≠ 0)
So, one stationary point occurs when x = 0.
4. Determining whether the value is a maximum or minimum:
To determine whether the value at x = 0 is a maximum or minimum, we need to find the second derivative of V with respect to x and evaluate it at x = 0.
Taking the derivative of dV/dx = 2x * y with respect to x gives us:
d²V/dx² = 2y
At x = 0, we have:
d²V/dx² = 2y ≠ 0
Since the second derivative is positive (2y > 0), we can conclude that the value at x = 0 is a minimum.
Therefore, the value of x is 0 and the value of y can vary for the volume V to have a stationary value (minimum).