To solve Question 7, where we need to find the dimensions of a rectangular box open at the top, of maximum capacity, whose surface area is 432 sq. cm, follow the steps below:

Step-by-Step Solution

1. Define variables:

• Let the length of the rectangular box be $l$,
• Let the width be $w$,
• Let the height be $h$.

Since the box is open at the top, the surface area includes:

• 1 base (area = $l \times w$),
• 4 sides (2 rectangles each of area $l \times h$ and $w \times h$).

Thus, the total surface area is: $S = lw + 2lh + 2wh$ Given that $S = 432$ sq. cm: lw + 2lh + 2wh = 432 \tag{1}

2. Objective: Maximize the volume

The volume $V$ of the box is given by: V = l \cdot w \cdot h \tag{2}

Our goal is to maximize $V$ under the constraint given by equation $(1)$.

3. Express $h$ in terms of $l$ and $w$:

From equation $(1)$: $2lh + 2wh = 432 - lw$ Factor out $h$: $h(2l + 2w) = 432 - lw$ Solve for $h$: h = \frac{432 - lw}{2(l + w)} \tag{3}

4. Substitute $h$ into the volume equation:

Substitute $h$ from equation $(3)$ into $V = l \cdot w \cdot h$: $V = l \cdot w \cdot \frac{432 - lw}{2(l + w)}$ Simplify: V = \frac{lw (432 - lw)}{2(l + w)} \tag{4}

5. Use partial derivatives to maximize $V$:

To find the maximum volume, differentiate $V$ with respect to $l$ and $w$, and set the partial derivatives equal to zero:

1. $\frac{\partial V}{\partial l} = 0$,
2. $\frac{\partial V}{\partial w} = 0$.

This step involves solving a system of nonlinear equations, which can be tedious by hand. Substituting specific values or using numerical methods simplifies the task.

6. Final answer:

After solving (by symmetry or numerical computation), the dimensions of the box that maximize the volume under the given constraint are:

• Length $l = 12$ cm,
• Width $w = 12$ cm,
• Height $h = 6$ cm.

Verification:

1. Surface area: $S = lw + 2lh + 2wh = 12 \cdot 12 + 2(12 \cdot 6) + 2(12 \cdot 6) = 432 \, \text{cm}^2$
2. Volume: $V = l \cdot w \cdot h = 12 \cdot 12 \cdot 6 = 864 \, \text{cm}^3.$

Thus, the solution satisfies the conditions.

Let me know if you want further clarification or detailed steps for the derivatives.

Related Questions:

1. How do you solve optimization problems using Lagrange multipliers?
2. What is the significance of partial derivatives in maxima and minima?
3. How can you generalize this problem to closed rectangular boxes?
4. What happens if the surface area constraint changes?
5. How do you verify that a critical point gives a maximum volume?

Tip:

In optimization problems, always express one variable in terms of others using the constraint equation to reduce the number of variables.