A rectangle is inscribed in a circle of radius 5 units centre O(0, 0) as shown below - Leaving Cert Mathematics - Question 8 - 2020

The area A(θ) of the rectangle can be expressed as:

$A(θ) = (a imes b) = (5 ext{cos}( heta)) imes (5 ext{sin}( heta)) = 25 ext{sin}(2 heta)$

Now using:

$ext{sin}(2 heta) = 2 ext{sin}( heta) ext{cos}( heta)$

Therefore, we have:

$A(θ) = 50 ext{sin}(2 heta).$

To find the maximum area, we need to differentiate A(θ):

$A'(θ) = 50 ext{cos}(2θ)$

Setting the derivative to zero:

50 ext{cos}(2θ) = 0 \\ ext{cos}(2θ) = 0$$$$\Rightarrow 2θ = \frac{ heta}{2} \\ θ = \frac{π}{4}

To confirm this is a maximum, we check the second derivative:

$A''(θ) = -100 ext{sin}(2θ)$

Evaluating $A''(θ)$ at $θ = \frac{π}{4}$ yields a negative value, confirming a maximum, which occurs when a = b, thus indicating a square.

Substituting $θ = \frac{π}{4}$ into A(θ):

$A(\frac{π}{4}) = 50 ext{sin}\left(2 \cdot \frac{π}{4}\right) = 50 ext{sin}(\frac{π}{2}) = 50$

Thus, the maximum area is:

$50 \text{ square units}$.

Let l be the distance from the person to the base of the streetlight. By similar triangles:

$\frac{5}{l} = \frac{2}{x} \\ \Rightarrow 5x = 2l \\ \Rightarrow l = \frac{5x}{2}$

Differentiating with respect to time:

$\frac{dl}{dt} = \frac{5}{2} \frac{dx}{dt}$

Given that the person is walking towards the streetlight at 1.5 m/s, we have:

$\frac{dl}{dt} = -1.5 m/s$

Substituting this value in:

-1.5 = \frac{5}{2} \frac{dx}{dt} \\ rac{dx}{dt} = -\frac{3}{5} = -0.6

Thus, the length of the person’s shadow is decreasing at a rate of 0.6 m/s.