Question

A rectangular closed block with a square base has the total surface area of 150 s cms. Find its dimensions which maximize the volume and also the maximum volume.

Answer 1

Dimensions: 5 cm x 5 cm x 5 cm, Maximum Volume: 125 cubic cms

Answer 2

To solve this problem, we need to find the dimensions of the rectangular closed block with a square base that maximize its volume, given its total surface area.

Let the side of the square base be $x$ cm and the height of the block be $h$ cm.

1. Formulate the total surface area ($A$) equation:
   A closed rectangular block with a square base has two square faces (top and bottom) and four rectangular side faces.
   Area of the base = $x^2$
   Area of the top = $x^2$
   Area of the four side faces = $4 \times (x \times h) = 4xh$
   Total surface area $A = x^2 + x^2 + 4xh = 2x^2 + 4xh$.
   Given $A = 150$ sq cms.
   So, $2x^2 + 4xh = 150$.

2. Express $h$ in terms of $x$:
   From the surface area equation:
   $4xh = 150 - 2x^2$
   $h = \frac{150 - 2x^2}{4x} = \frac{2(75 - x^2)}{4x} = \frac{75 - x^2}{2x}$.

3. Formulate the volume ($V$) equation:
   The volume of the block is given by $V = \text{Area of base} \times \text{height}$.
   $V = x^2 \times h$
   Substitute the expression for $h$ into the volume equation:
   $V(x) = x^2 \left( \frac{75 - x^2}{2x} \right)$
   $V(x) = \frac{x(75 - x^2)}{2}$
   $V(x) = \frac{75x - x^3}{2}$.

4. Find the critical points by differentiating $V(x)$ with respect to $x$ and setting the derivative to zero:
   $V'(x) = \frac{d}{dx} \left( \frac{75x - x^3}{2} \right)$
   $V'(x) = \frac{1}{2} (75 - 3x^2)$.
   To find the maximum volume, set $V'(x) = 0$:
   $\frac{1}{2} (75 - 3x^2) = 0$
   $75 - 3x^2 = 0$
   $3x^2 = 75$
   $x^2 = 25$
   $x = \pm 5$.
   Since $x$ represents a dimension, it must be positive. Therefore, $x = 5$ cms.

5. Use the second derivative test to confirm if it's a maximum:
   Find the second derivative of $V(x)$:
   $V''(x) = \frac{d}{dx} \left( \frac{1}{2} (75 - 3x^2) \right)$
   $V''(x) = \frac{1}{2} (0 - 6x)$
   $V''(x) = -3x$.
   Now, evaluate $V''(x)$ at $x = 5$:
   $V''(5) = -3(5) = -15$.
   Since $V''(5) < 0$, the volume is indeed maximized at $x = 5$ cms.

6. Calculate the height ($h$) and the maximum volume:
   Substitute $x = 5$ cms back into the equation for $h$:
   $h = \frac{75 - x^2}{2x} = \frac{75 - (5)^2}{2(5)}$
   $h = \frac{75 - 25}{10} = \frac{50}{10}$
   $h = 5$ cms.
   So, the dimensions that maximize the volume are 5 cm (length of base) $\times$ 5 cm (width of base) $\times$ 5 cm (height). This means the block is a cube.

   Now, calculate the maximum volume:
   $V_{max} = x^2 h = (5)^2 \times 5 = 25 \times 5 = 125$ cubic cms.

The dimensions which maximize the volume are 5 cm $\times$ 5 cm $\times$ 5 cm, and the maximum volume is 125 cubic cms.

Explanation of the solution:

1. Define variables for dimensions (x for base side, h for height).
2. Write the total surface area formula for a closed block with a square base: $A = 2x^2 + 4xh$.
3. Substitute the given surface area $A=150$ and express $h$ in terms of $x$: $h = \frac{75 - x^2}{2x}$.
4. Write the volume formula: $V = x^2h$.
5. Substitute the expression for $h$ into the volume formula to get $V(x) = \frac{75x - x^3}{2}$.
6. Differentiate $V(x)$ with respect to $x$: $V'(x) = \frac{1}{2}(75 - 3x^2)$.
7. Set $V'(x) = 0$ and solve for $x$: $x=5$ (since dimension must be positive).
8. Calculate the second derivative $V''(x) = -3x$.
9. Evaluate $V''(5) = -15 < 0$, confirming that $x=5$ corresponds to a maximum volume.
10. Substitute $x=5$ back into the expression for $h$ to find $h=5$ cm.
11. Calculate the maximum volume using $x=5$ and $h=5$: $V_{max} = 5^2 \times 5 = 125$ cubic cms.