A rectangular sheet of metal with dimensions 20 cm by 15 cm is to be used to create an open top box by cutting a square, x cm by x cm, from each corner and bending up the sides. If a volume of 375 cm^3 is required, determine the side length of the squares that must be cut.

We are tasked with determining the side length \( x \) of the squares that must be cut from each corner of a rectangular sheet of metal to create an open-top box with a given volume of 375 cm³.

### Step-by-step approach:

1. **Dimensions of the base of the box:**
   - After cutting out squares from each corner, the new dimensions of the base will be reduced by \( 2x \) on both sides.
   - The new length of the box's base: \( 20 - 2x \) cm.
   - The new width of the box's base: \( 15 - 2x \) cm.
   
2. **Height of the box:**
   - After cutting the squares, the height of the box will be equal to the side length \( x \) of the squares.

3. **Volume of the box:**
   The volume of the box is given by the formula:
   \[
   V = \text{length} \times \text{width} \times \text{height}
   \]
   Substituting the dimensions:
   \[
   375 = (20 - 2x)(15 - 2x)(x)
   \]
   This is the equation we need to solve for \( x \).

### Step 1: Expand the equation

First, expand \( (20 - 2x)(15 - 2x) \):
\[
(20 - 2x)(15 - 2x) = 300 - 40x - 30x + 4x^2 = 300 - 70x + 4x^2
\]

Now, substitute this expression back into the volume equation:
\[
375 = (300 - 70x + 4x^2)(x)
\]
\[
375 = 300x - 70x^2 + 4x^3
\]

### Step 2: Rearrange the equation

Rearrange the equation to set it to 0:
\[
4x^3 - 70x^2 + 300x - 375 = 0
\]

### Step 3: Solve the cubic equation

We can solve the cubic equation using numerical methods, factoring, or graphing tools to find the value of \( x \). In this case, the cubic equation simplifies numerically to give a solution for \( x \).

Using numerical approximation methods or factoring techniques, we find that the solution is approximately:
\[
x \approx 2.5 \, \text{cm}
\]

### Conclusion:
The side length of the squares that must be cut from each corner is approximately \( \boxed{2.5} \, \text{cm} \).

Would you like more details on the solving process or further explanation?

### Related Questions:
1. What would happen to the volume if the side length of the square \( x \) was smaller?
2. How would the volume change if the original dimensions of the sheet were different?
3. Can you calculate the surface area of the open-top box after cutting and folding?
4. What is the maximum possible volume for this box given the same sheet dimensions?
5. How would the process change if the box had a top?

**Tip:** When solving for \( x \), it’s helpful to factor or use numerical methods for cubic equations, especially if exact factoring is difficult.

Learn how to calculate the side length of squares to cut from each corner of a rectangular sheet of metal to create an open-top box with a volume of 375 cm³. This problem involves algebraic manipulation, cubic equations, and volume calculations, suitable for students in Grades 10-12.

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