A carpenter wants to sell 40 chairs. I f he sells them at ₹ 156 per chair, he would be able to sell all the chairs. But for every ₹ 6 increase in price, he will be left with one additional unsold chair. At what selling price would he be able to maximize his profits (assuming unsold chairs remain with him)?
(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 - > 156 × 40 84x - 6 > 0 Differentiate the above equation and putting = 0 84 -12x = 0 ⇒ 84 = 12x ∴ x = 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