Answer :

Given triangle:

Let Base PQ = a

Hypotenuse PR = b

Let θ be the angle between them.

Let a + b = k (constant)

∴ b = k – a … (1)

In right angled Δ PQR,

⇒ QR^{2} = PR^{2} – PQ^{2}

= b^{2} – a^{2}

= (k – a)^{2} – a^{2}

= k^{2} + a^{2} – 2ka – a^{2}

∴ QR = k^{2} – 2ka … (2)

Area of ΔPQR:

⇒ A = 1/2 × PQ × QR

⇒ A^{2} = 1/4 × a^{2} × QR^{2}

From (2),

∴ A^{2} = 1/4 × a^{2} × (k^{2} – 2ka)

Let A^{2} = Z then Z = 1/4 × a^{2} × (k^{2} – 2ka)

Differentiating both sides w. r. to a, we get

… (3)

For maxima or minima dZ/da = 0

⇒ ka/2 = 0 or k – 3a = 0

∴ a = 0 or a = k/ 3

Again differentiating (3) on both sides w. r. to a, we get

∴ Z is maximum at a = k/3.

A^{2} is maximum at k/3, and A is maximum at a = k/3

From (1),

∴ θ = π/3

∴ Area of Δ is maximum when the angle between base and hypotenuse is π/3.

**OR**

Let S (x) and C (x) be the selling and cost price of x items respectively,

And

We know that the profit function p (x) is Given: by p (x) = S (x) – C ( x)

… (1)

… (2)

We know that for maximum profit, p’(x) = 0

∴ x = 240

Now

∴ Profit is maximum when x = 240 items.

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