MAT Exam 2020 Paper

Concept:The carpenter maximizes revenue when the increase in price reduces the number of chairs sold optimally.Explanation:Let the price be increased $K$ times by ₹6 each time.Then the new selling price is $156 + 6K$ rupees per chair.The number of chairs sold becomes $40 - K$.Total revenue is:$$(156 + 6K)(40 - K)$$For maximum revenue, this must be greater than the original revenue $156 \times 40$:$$(156 + 6K)(40 - K) > 156 \times 40$$Expanding:$$156 \times 40 + 240K - 156K - 6K^2 > 156 \times 40$$Simplifying:$$84K - 6K^2 > 0$$To find the maximum, differentiate with respect to $K$ and set equal to zero:$$\frac{d}{dK}(84K - 6K^2) = 84 - 12K = 0$$$$84 - 12K = 0 \Rightarrow K = 7$$Second derivative:$$\frac{d}{dK}(84 - 12K) = -12 < 0$$Since the second derivative is negative, $K = 7$ gives maximum revenue.Maximum selling price:$$156 + 6K = 156 + 6(7) = 156 + 42 = ₹198$$Answer:₹198 (Option A)