A parcel in the form of a rectangular block is held together by three pieces of tape as shown in the diagram . 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??

Oh, don't worry, I'll break it down for you! So, the parcel is like a rectangular block, right? And it's held together by three pieces of tape. Now, let's talk math.

To find the total length of tape, we need to consider that there are two ends of the parcel, which are squares of side x cm. So, the total length of tape needed for both ends is 4x cm.

Now, the parcel has a length of y cm. So, the total length of tape needed to go around the sides of the parcel is 2y cm.

Adding up both lengths of tape, we get the total length of tape as 4x + 2y cm.

Given that the total length of tape is 450 cm, we can set up an equation:
4x + 2y = 450

Now, let's talk about the volume of the parcel, V cm3. The volume of a rectangular block is given by the formula:
V = x^2 * y

To find the value of x and y for which V has a stationary value (either a maximum or a minimum), we need to find the critical points of the function V = x^2 * y. To do that, we can take partial derivatives of V with respect to x and y and set them equal to zero.

Once we find those critical points, we can determine whether they correspond to maximum or minimum values by analyzing the second derivative of V.

So, now it's on you to solve for x and y! Good luck, my friend!

This problem is asking for the total length of tape used in a rectangular parcel and the volume of the parcel, both in terms of x and y. It also asks for the value of x and y that will yield a stationary value for the volume, and whether this value is a maximum or minimum.

To begin, let's analyze the diagram and understand the structure of the rectangular parcel. The parcel has square ends with side lengths of x cm, and its length is y cm. The three pieces of tape wrap around the parcel, holding it together.

Now, let's calculate the total length of tape used. There are two pieces of tape wrapping around the width of the parcel, and one piece wrapping around the length. The length of the tape wrapping around the width is equal to the perimeter of the square end, which is 4 times the side length, 4x cm. The length of the tape wrapping around the length is simply y cm. Therefore, the total length of tape used is 4x + y cm.

Given that the total length of tape used is 450 cm, we can write the equation: 4x + y = 450.

Next, let's express the volume, V, of the parcel in terms of x. The volume of a rectangular parcel is given by the formula V = length x width x height. In this case, the length is y cm, the width is x cm, and the height can be determined by looking at the diagram. The height appears to be the same as the width, which is x cm. Therefore, the volume, V, of the parcel is V = y * x * x = x^2 * y cm^3.

To find the value of x and y for which V has a stationary value, we need to differentiate the volume equation with respect to x and set it equal to zero. Let's do that:

dV/dx = 2x * y = 0.

To find the value of x, we need to solve the equation 2x * y = 0. Since y is not equal to zero (otherwise, we would have a flat parcel with no volume), we can conclude that x must be equal to zero. However, a parcel with sides of length zero is not possible, so there is no real solution for x.

Therefore, there is no stationary value for the volume, V, and we cannot determine if it is a maximum or minimum.

Let's break down the problem step by step:

1. Total Length of Tape:
The parcel consists of three pieces of tape. Two pieces are used to hold the ends of the parcel together, and one piece is used to hold the length of the parcel. Since the ends of the parcel are square with side length x cm, the length of tape used for each end is 4 times the side length, which is 4x cm. The tape used for the length of the parcel is y cm. Therefore, the total length of tape used is 2(4x) + 1(y) = 8x + y cm.

2. Volume of the Parcel:
To find the volume of the parcel, we need to multiply the length, width, and height. The length is given as y cm. The width and height are both equal to x cm since the ends of the parcel are square. Therefore, the volume V can be expressed as V = y * x * x = xy^2 cm^3.

3. Stationary Value of V:
To find the stationary value of V, we need to determine the values of x and y where the derivative of V with respect to x is zero. First, let's differentiate V with respect to x:

dV/dx = y * 2x = 2xy

Setting the derivative equal to zero:

2xy = 0

Since y is not zero (given that the parcel has positive dimensions), the only way for the product 2xy to be zero is if x equals zero. However, x cannot be zero because it represents the side length of the ends of the parcel. Therefore, there is no stationary value for V.

It is important to note that without additional information, such as constraints on the values of x and y, we cannot determine whether V would reach a maximum or minimum at a stationary point.