The diagonal of a square, of side $3\sqrt{2}$ cm, is increasing at a rate of 2 cm/s. Which of the following is the rate at which its area is increasing?

$12 \text{ cm}^2/\text{s}$

The correct answer is Option (3) → $12 \text{ cm}^2/\text{s}$ ##

By the Pythagorean theorem:

$D = s\sqrt{2} $

Thus, the side $s$ in terms of the diagonal $D$:

$s = \frac{D}{\sqrt{2}}$

The area $A$ of the square is given by:

$A = s^2 $

Substitute $s = \frac{D}{\sqrt{2}}$ into the formula for $A$:

$A = \left( \frac{D}{\sqrt{2}} \right)^2 = \frac{D^2}{2} $

To find the rate of change of the area, $\frac{dA}{dt}$, we differentiate both sides of the area formula with respect to time $t$:

$\frac{dA}{dt} = \frac{d}{dt} \left( \frac{D^2}{2} \right) $

$\frac{dA}{dt} = \frac{2D}{2} \cdot \frac{dD}{dt} = D \cdot \frac{dD}{dt}$

The diagonal $D$ of the square when $s = 3\sqrt{2}$ cm can be calculated as:

$ D = s\sqrt{2} = 3\sqrt{2} \cdot \sqrt{2} = 3 \times 2 = 6 \text{ cm} $

Now, substitute $D = 6 \text{ cm}$ and $\frac{dD}{dt} = 2 \text{ cm/s}$ into the derivative formula:

$ \frac{dA}{dt} = 6 \times 2 = 12 \text{ cm}^2/\text{s} $