To find the largest possible volume of the box with a square base and an open top, we need to use optimization techniques. Here's how we approach it:

Step 1: Define the variables

Let:

• $x$ be the length of the side of the square base (in cm),
• $h$ be the height of the box (in cm).

Step 2: Write the surface area equation

The surface area $A$ consists of:

• The area of the square base: $x^2$,
• The area of the four sides (each with area $xh$): $4xh$.

The total surface area is given by the sum of these areas, and we know that the total surface area is 1100 cm²: $x^2 + 4xh = 1100.$

Step 3: Write the volume equation

The volume $V$ of the box is: $V = x^2 h.$

Step 4: Solve for $h$ in terms of $x$

From the surface area equation: $x^2 + 4xh = 1100,$ solve for $h$: $4xh = 1100 - x^2,$ $h = \frac{1100 - x^2}{4x}.$

Step 5: Substitute $h$ into the volume equation

Now, substitute this expression for $h$ into the volume equation $V = x^2 h$: $V = x^2 \left( \frac{1100 - x^2}{4x} \right),$ $V = \frac{x(1100 - x^2)}{4}.$

Step 6: Simplify the volume equation

Simplify the equation for volume: $V = \frac{1100x - x^3}{4}.$

Step 7: Maximize the volume

To find the maximum volume, we need to differentiate $V$ with respect to $x$, and then set the derivative equal to zero to find the critical points.

First, differentiate: $\frac{dV}{dx} = \frac{1100 - 3x^2}{4}.$

Set $\frac{dV}{dx} = 0$ to find critical points: $1100 - 3x^2 = 0,$ $3x^2 = 1100,$ $x^2 = \frac{1100}{3},$ $x = \sqrt{\frac{1100}{3}} \approx 19.17 \, \text{cm}.$

Step 8: Verify if this gives a maximum

To verify if this is a maximum, we check the second derivative: $\frac{d^2V}{dx^2} = -\frac{6x}{4} = -\frac{3x}{2}.$ Since $x > 0$, $\frac{d^2V}{dx^2} < 0$, confirming that this critical point corresponds to a maximum.

Step 9: Calculate the maximum volume

Now, substitute $x = 19.17$ cm into the equation for $h$: $h = \frac{1100 - (19.17)^2}{4(19.17)} \approx \frac{1100 - 367.5}{76.68} \approx \frac{732.5}{76.68} \approx 9.55 \, \text{cm}.$

Finally, the maximum volume is: $V = x^2 h = (19.17)^2 \times 9.55 \approx 367.5 \times 9.55 \approx 3503.63 \, \text{cm}^3.$

Final Answer:

The largest possible volume of the box is approximately 3503.63 cubic centimeters.

Would you like further details on the solution or have any specific questions?

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Here are 5 related questions:

1. How do you optimize a function to find maximum or minimum values?
2. What are the steps to solve optimization problems involving constraints?
3. How can you apply these methods to a box with a different shape or constraints?
4. How would the solution change if the box had a top as well as a bottom?
5. How can you check whether your solution gives a maximum or a minimum?

Tip: When solving optimization problems, always check if the second derivative is negative at the critical points to confirm a maximum.