If the selling price on any given day is the same for all the plates and can only be a multiple of Rs. 10, then what is the maximum pro t that FS can achieve in a day?
C
Step-by-Step Solution
Key idea: This is a revenue optimization problem with a linear demand curve and discrete constraints. We need to maximize Profit = (Selling Price - Cost Price) * Quantity Sold.
Step 1: Define the demand function.
Base case: Price , Quantity .
Change: For every Rs. 10 increase in P, Q decreases by 10.
Let be the number of Rs. 10 increases.
.
.
Constraints:
- .
- (Capacity).
- is a multiple of 10. (Satisfied by formula).
- .
Step 2: Formulate Profit function.
Factor out 10:
Step 3: Maximize the quadratic function.
Let .
Roots are at and .
The vertex (maximum) of a parabola is at the midpoint of roots.
.
Check if is valid.
. (, OK).
. (, OK).
Step 4: Calculate Max Profit.
.
.
Wait, let me double check the capacity constraint.
"More than 400 plates cannot be produced".
Our Q is 170, which is well within 400.
Is it possible to decrease price?
If is negative.
Let (decreases).
.
.
Constraint .
Profit .
.
Roots at .
Vertex at .
Since parabola opens downward, max is at vertex. But domain is .
Vertex -13 is outside. Function decreases as we move away from -13?
Actually, vertex is at -13. For , the function is decreasing?
Let's check derivative or values.
.
This is decreasing for all .
So max profit in this region is at (no change).
.
Comparing regions:
Increase price (): Max 28,900 at .
Decrease price (): Max 12,000 at .
Global Max is 28,900.
Let me re-read the options.
A: 25,300
B: 28,900
C: 41,400
D: 52,900
My calculation gives 28,900. This matches Option B.
Wait, did I miss something?
"FS incurs a cost of Rs. 120 per plate".
"Selling price ... multiple of Rs. 10".
Let's re-calculate .
.
.
.
.
Why is Option C 41,400?
Maybe I made a mistake in the demand slope?
"For each increase ... by 10 ... 10 fewer". Slope is -1.
.
.
Profit .
Vertex at .
.
Profit .
Calculation is robust.
Let's check if capacity constraint binds in the other direction?
If we decrease price, Q increases.
Max Q is 400.
.
But .
So min P is 120.
At , .
Profit .
So the range of P is .
Vertex 290 is in range.
Answer is 28,900.
Option B is 28,900.
Why did I write C in the header? I must correct it.
Answer: Option B.