Question

How many units should Mr. David produce daily?

• A. 130
• B. 100
• C. 70
• D. 150
• E. Cannot be determined
• A. 620
• B. 920
• C. 840
• D. 760
• J. Cannot be determined

Answer 1

The Correct Option is B

Given: Cost function \( C(x) = 240 + bx + cx^2 \). From data: When production increases from 20 to 40, cost increases by \( \frac{2}{3} \) of original cost at 20 units: \[ C(40) - C(20) = \frac{2}{3} C(20). \] Similarly, from 40 to 60 units, cost increases by 50% of cost at 40 units: \[ C(60) - C(40) = 0.5 \ C(40). \] Substitute \( C(x) \): 1st equation: \[ [240 + 40b + 1600c] - [240 + 20b + 400c] = \frac{2}{3} [240 + 20b + 400c]. \] Simplify: \[ 20b + 1200c = \frac{2}{3} (240 + 20b + 400c). \] 2nd equation: \[ [240 + 60b + 3600c] - [240 + 40b + 1600c] = 0.5 [240 + 40b + 1600c]. \] Simplify: \[ 20b + 2000c = 0.5 (240 + 40b + 1600c). \] From first: Multiply through by 3: \( 60b + 3600c = 480 + 40b + 800c \) → \( 20b + 2800c = 480 \) → \( 5b + 700c = 120 \) → (i).
From second: \( 20b + 2000c = 120 + 20b + 800c \) → \( 1200c = 120 \) → \( c = 0.1 \).
From (i): \( 5b + 700(0.1) = 120 \) → \( 5b + 70 = 120 \) → \( 5b = 50 \) → \( b = 10 \).
Thus \( C(x) = 240 + 10x + 0.1x^2 \).
Revenue: \( R(x) = 30x \).
Profit: \( P(x) = 30x - (240 + 10x + 0.1x^2) = -0.1x^2 + 20x - 240 \).
Maximise profit: \( \frac{dP}{dx} = -0.2x + 20 = 0 \) → \( x = 100 \).

Answer 2

At \( x = 100 \): Revenue = \( R(100) = 30 \times 100 = 3000 \).
Cost = \( C(100) = 240 + 10 \times 100 + 0.1 \times 10000 = 240 + 1000 + 1000 = 2240 \).
Profit = \( 3000 - 2240 = 760 \) wait — recheck.
Rechecking derivative point: \( P(x) = -0.1x^2 + 20x - 240 \).
At \( x = 100 \): \( P(100) = -0.1(10000) + 2000 - 240 = -1000 + 2000 - 240 = 760 \).
Thus Correct Answer is actually (4) 760.