Chemistry Daily Challenge 26-July-2015

Chemistry Level 3

A chemist wants to start small business by selling B.

B is produced from A by following reaction A ⇌ B A \rightleftharpoons B Final conversion of A is 75%.Assume leftover A can't be reused. For this business, he requires to buy A and other equipments Before he can start.He wants to know about profit.Let N be kg of product B formed in one day.

Selling Price \color{#624F41}{\text{Selling Price}}

1) Cost of B=100 Rupee/kg

Expenditures \color{#624F41}{\text{Expenditures}}

1) Cost of A=50 Rupee/kg

2) All other expenditure= 1000 + N 2 30 1000+\frac{N^2}{30} Rupee/day

Help him finding Maximum profit he can make in one day. \color{#3D99F6}{\text{Help him finding Maximum profit he can make in one day.}}

Hint: Profit=(Total Selling price)-(Total expenditure)


The answer is 7333.333.

Aamir Faisal Ansari by Aamir Faisal Ansari , 24, India

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2 solutions

Profit=Selling price-Cost price

From reaction,one kg of A gives one kg of B.To produce N kg of B,we must use 4 3 N \frac{4}{3}N kg A.Since left over A that is 1 3 N \frac{1}{3}N can't be reused.

Profit= 100 N − ( 50 × 4 3 × N + 1000 + N 2 30 ) 100N-(50\times \frac{4}{3}\times N+1000+\frac{N^2}{30})

⟹ 100 N 3 − N 2 30 − 1000 \implies \frac{100N}{3}-\frac{N^2}{30}-1000

For maximizing profit, d ( P r o f i t ) d N = 0 \frac{d(Profit)}{dN}=0

⟹ 100 3 − 2 N 30 = 0 \implies \frac{100}{3}-\frac{2N}{30}=0

⟹ N = 500 \implies N=500

Hence

Profit= 100 × 500 3 − 50 0 2 30 − 1000 = 22000 3 = 7333.333333 \frac{100\times 500}{3}-\frac{500^2}{30}-1000=\frac{22000}{3}=7333.333333

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Did the same way. Nice solution! Upvoted!

Nelson Mandela - 5 years, 10 months ago

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Thanks dear.........Your words are really encouraging.

Aamir Faisal Ansari - 5 years, 10 months ago

Did it the same way!Upvoted!

Athiyaman Nallathambi - 5 years, 10 months ago

Mass of A A used per day is 4 N 3 \dfrac{4N}{3} kg. Then, the total expenditure is

Rs. 1000 + 200 N 3 + N 2 30 1000+\dfrac{200N}{3}+\dfrac{N^2}{30} .

Total money earned is

Rs. 100 N 100N per day.

Hence, profit per day is

Rs. 100 N − 200 N 3 − N 2 30 − 1000 100N-\dfrac{200N}{3}-\dfrac{N^2}{30}-1000

= = Rs. ( − 1 30 ( N 2 − 1000 N ) − 1000 ) \left (-\dfrac{1}{30}(N^2-1000N)-1000\right )

= = Rs. ( − 1 30 ( N − 500 ) 2 + 25000 3 − 1000 ) ≤ \left (-\dfrac{1}{30}(N-500)^2+\dfrac{25000}{3}-1000\right )\leq Rs. 22000 3 ≈ \dfrac{22000}{3}\approx Rs. 7333.3333 7333.3333 .

Hence the maximum profit per day is Rs. 7333.3333 \boxed {7333.3333} .

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