Question:

A rectangular sheet of metal measuring 10 inches by 15 inches is used to create an open box by cutting equal squares from each corner and folding up the sides. Determine the dimensions of the squares cut from the corners that will maximize the volume of the box. Show all work and justify your answer.

Answer:

Let's start by assigning variables to the unknowns in the problem:

Let the length of each side of the square cut from the corners be denoted by $x$ inches.

To find the dimensions of the squares cut from the corners that will maximize the volume of the box, we need to express the volume of the box in terms of $x$ and then find the value of $x$ that maximizes the volume.

We also need to consider the restrictions on possible values for $x$. Since we are cutting squares from each corner of the rectangular sheet, the length of each side of the square ($x$) must be less than half the length of the corresponding side of the rectangle. Therefore, we have the following inequality:

Now, let's proceed to find the volume of the box:

The length, width, and height of the box will be given by the following expressions:

Length of the box: $10 - 2x$ inches
Width of the box: $15 - 2x$ inches
Height of the box: $x$ inches

The volume, $V$, of the box is given by the product of its length, width, and height:

We can simplify this expression by multiplying the factors:

To maximize the volume, we need to find the value of $x$ that maximizes the cubic function $V(x)$.

To do this, we can take the derivative of $V$ with respect to $x$:

Setting $V'(x)$ equal to zero, we can solve for $x$:

Factoring this quadratic equation, we get:

Solving for $x$, we have two possible solutions:

$2x - 5 = 0 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2} = 2.5$

$6x - 30 = 0 \Rightarrow 6x = 30 \Rightarrow x = 5$

However, since the value of $x$ must be strictly less than 5 (as stated by the restrictions on $x$), we discard the solution $x = 5$.

Now, let's determine whether the critical point at $x = \frac{5}{2}$ corresponds to a maximum or minimum.

To do this, we can examine the concavity of the function $V(x)$ by taking the second derivative:

Substituting $x = \frac{5}{2}$ into $V''(x)$, we have:

Since $V''\left(\frac{5}{2}\right) < 0$, the concavity is negative at $x = \frac{5}{2}$. This indicates that the critical point corresponds to a maximum value of the volume function.

Therefore, the dimensions of the squares cut from the corners that will maximize the volume of the box are $x = \frac{5}{2}$ inches.

To find the actual dimensions of the box, we substitute this value of $x$ back into the expressions for the length, width, and height:

Length of the box: $10 - 2x = 10 - 2\left(\frac{5}{2}\right) = 5$ inches
Width of the box: $15 - 2x = 15 - 2\left(\frac{5}{2}\right) = 10$ inches
Height of the box: $x = \frac{5}{2}$ inches

Thus, the dimensions of the box that maximizes the volume are 5 inches by 10 inches by $\frac{5}{2}$ inches.

The maximum volume, $V$, of the box can be calculated using the volume formula:

[V = (5)(10)\left(\frac{5}{2}\right) = 25 \times 5 = 125) cubic inches.

Therefore, the dimensions of the squares cut from the corners that will maximize the volume of the box are $x = \frac{5}{2}$ inches and the maximum volume is 125 cubic inches.