A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is conver

Q: Wait, why do we subtract $2x$ from the length and width? Shouldn't it just be $x$?

That is a classic trap! Imagine the sheet. You cut a square of side $x$ from the left corner AND the right corner. That is $x + x = 2x$ removed from the total length. Always visualize the fold—the flap you lift up is $x$, but the base of the box is what remains between the cuts.

Q: Why do we reject the root $k = 5/6$?

Great eye for detail! If $k = 5/6$, the width $8k$ becomes $8(5/6) = 6.67$. But we removed $2x = 10$ from the width. You cannot remove $10$ units from a $6.67$ unit side. It is physically impossible, so we must discard it.

Q: Why is $V'(x) = 0$ the maximum? Could it be a minimum?

Excellent skepticism. In this physical context, the volume is zero at the boundaries (where $x=0$ or where the sheet is fully consumed). Since the volume is positive between these bounds, the critical point we find must be a local maximum. You are thinking like a true mathematician!

Question

A rectangular sheet of fixed perimeter with sides having their lengths in the ratio $8 : 15$ is converted into an open rectangular box by folding after removing squares of equal area from all four corners. If the total area of removed squares is $100$, the resulting box has maximum volume. Then the lengths of the sides of the rectangular sheet are

Options:

(A) 24

(B) 32

(C) 45

(D) 60

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a workshop. In front of you lies a rectangular sheet of paper. It is not just a piece of paper; it is a canvas of potential.

We are told the sides are in a ratio of $8:15$. Let us define these sides as $8k$ and $15k$. This constant $k$ is our scaling factor, the hidden DNA of our rectangle.

### The Act of Removal

To create this box, we must perform a surgical operation. We remove squares of equal area from all four corners. Let the side of each square be $x$.

We are given that the total area of these four squares is $100$. Mathematically, this is $4x^2 = 100$, which simplifies beautifully to $x^2 = 25$, giving us $x = 5$.

We now know exactly how much we are cutting away. This is the first step in our journey: defining the constraints of our physical reality.

### The Algebra of Folding

Now, visualize the folding process. When you fold up the flaps, the height of your box becomes $x$.

The original length was $15k$. By cutting a square of side $x$ from both ends, we have reduced the length by $2x$. Thus, the new length is $15k - 2x$.

Similarly, the width becomes $8k - 2x$. The volume $V$ is the product of these three dimensions:
$$V(x) = (15k - 2x)(8k - 2x)x$$

### The Calculus of Optimization

We want to maximize this volume. In the language of calculus, this means we need to find the point where the rate of change of volume with respect to the cutout size is zero.

Let us expand our volume function:
$$V(x) = (120k^2 - 30kx - 16kx + 4x^2)x$$
$$V(x) = 4x^3 - 46kx^2 + 120k^2x$$

Now, we differentiate with respect to $x$:
$$\frac{dV}{dx} = 12x^2 - 92kx + 120k^2$$

We know that the maximum volume occurs at $x = 5$. Therefore, we set $\frac{dV}{dx} = 0$ at $x = 5$.

### The Reality Check

Substituting $x = 5$ into our derivative, we get:
$$12(25) - 92k(5) + 120k^2 = 0$$

This simplifies to $300 - 460k + 120k^2 = 0$. Dividing by $20$, we arrive at the elegant quadratic equation:
$$6k^2 - 23k + 15 = 0$$

Factoring this, we find $(6k - 5)(k - 3) = 0$. This gives us two potential values for $k$: $k = \frac{5}{6}$ or $k = 3$.

Here is where the physics meets the math. If $k = \frac{5}{6}$, the width $8k$ is $\frac{40}{6} \approx 6.67$.

Since we removed $2x = 10$ from the width, we cannot remove $10$ units from a $6.67$ unit side. Thus, we reject $k = \frac{5}{6}$ as physically impossible.

We are left with **$k = 3$**. The dimensions of our sheet are $8(3) = 24$ and $15(3) = 45$. We have successfully navigated the geometry, the algebra, and the physical constraints to find our answer.

Select Answer:

Visualized Solution

Visualizing the Sheet

Removing the Corners

Finding the Square Side $x$

Dimensions of the Folded Box

Formulating the Volume Function

Expanding the Volume Expression

Condition for Maximum Volume

Differentiating $V (x)$

Substituting $x = 5$

Simplifying the Equation

Solving for $k$

Validating the Roots

Final Calculation of Sides

The Sigma Insight: Maxima and Minima

Analyzing the Setup

The Act of Removal

The Algebra of Folding

The Calculus of Optimization

The Reality Check

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