Step-by-Step Solution

Topic: Maxima/Minima & Related Rates

(I) EXPRESSIONS FOR y AND AREA A

Using Pythagoras theorem in the right triangle:

\( x^2 + y^2 = h^2 \)   ⇒   \( y^2 = h^2 – x^2 \)

Area of a right triangle \( A = \frac{1}{2} \times \text{base} \times \text{height} \):

(II) DERIVATIVE AND CRITICAL POINT

To simplify differentiation, let us maximize \( Z = A^2 \) instead of \( A \).

\( Z = \frac{1}{4} x^2 (h^2 – x^2) = \frac{1}{4} (h^2 x^2 – x^4) \)

Differentiate Z w.r.t x:

Set derivative to zero for critical points:

\( 2h^2 x – 4x^3 = 0 \)   ⇒   \( 2x(h^2 – 2x^2) = 0 \)

Since \( x \neq 0 \), we have \( h^2 = 2x^2 \).

(III)(A) CONFIRMING MAXIMUM AREA

Find the second derivative of \( Z \):

\( \frac{d^2Z}{dx^2} = \frac{1}{4} (2h^2 – 12x^2) \)

Substitute \( h^2 = 2x^2 \) (from critical point):

\( \frac{d^2Z}{dx^2} = \frac{1}{4} (2(2x^2) – 12x^2) = \frac{1}{4} (-8x^2) = -2x^2 \)

Since \( -2x^2 < 0 \), the function is concave down.

(III)(B) RELATED RATES (Ladder Pulling)

Given: \( h = 5 \)m, \( y = 3 \)m, Rate \( \frac{dy}{dt} = -2 \) m/s (decreasing).

Step 1: Find x

\( x^2 + y^2 = 5^2 \)   ⇒   \( x^2 + 3^2 = 25 \)   ⇒   \( x = 4 \) m.

Step 2: Differentiate w.r.t time (t)

\( x^2 + y^2 = 25 \)

\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)   ⇒   \( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \)

Step 3: Substitute values

\( 4(\frac{dx}{dt}) + 3(-2) = 0 \)

\( 4(\frac{dx}{dt}) – 6 = 0 \)   ⇒   \( 4(\frac{dx}{dt}) = 6 \)