(a ) Let the carpenter increases the price x times,
then the total increase in price would be ₹ 6x.
Also, the chair sold at increased price = (40 - x)
To maximize the profit the selling price of chair with increase price must be greater than original selling price, ∴ (156+ 6x)(40-x) > 156 × 40
156 x 40 + 240x - 156x -
x2 > 156 × 40
84x - 6
x2 > 0
Differentiate the above equation and putting = 0
84 -12x = 0 ⇒ 84 = 12x
∴ x =
=7 To check, if value of x will maximize or not we will against differentiate the above equation and put the value of x and if the result is negative, then the profit is maximized and if positive, then profit is minimum So, on differentiating 84 -12x = -12
So, the result is negative, hence the profit is maximum. The maximum selling price = 156 + 6 x 7
=156 + 42 = ₹ 198