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)>156×40 156×40+240x−156x−6x2>156×40 84x−6x2>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+6×7 =156+42=‌ ₹ ‌198