Q. 295.0( 2 Votes )

If the sum of the

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,


QR2 = PR2 – PQ2


= b2 – a2


= (k – a)2 – a2


= k2 + a2 – 2ka – a2


QR = k2 – 2ka … (2)


Area of ΔPQR:


A = 1/2 × PQ × QR


A2 = 1/4 × a2 × QR2


From (2),


A2 = 1/4 × a2 × (k2 – 2ka)


Let A2 = Z then Z = 1/4 × a2 × (k2 – 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.


A2 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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