Let cost function (TC) = 60 + 7Q^2 and revenue function (TR) = 120Q 5Q^2 . Find out equilibrium level of output and profit. [3+2]

Solution

1. Equilibrium Level of Output (Q)

The equilibrium level of output is determined where Total Revenue (TR) equals Total Cost (TC).

Given:

• Total Cost (TC) = $60+7Q^2$
• Total Revenue (TR) = $120Q-5Q^2$

Set TR = TC to find equilibrium output: $$120Q-5Q^2=60+7Q^2$$ Bring all terms to one side: $$120Q-5Q^2-7Q^2-60=0$$ $$-12Q^2+120Q-60=0$$ Multiply by -1 to simplify: $$12Q^2-120Q+60=0$$ Divide by 12 to simplify further: $$Q^2-10Q+5=0$$

Solve the quadratic equation using the quadratic formula: $$Q=\frac{10\pm \sqrt{(-10{)}^2-4\cdot 1\cdot 5}}{2\cdot 1}$$ $$Q=\frac{10\pm \sqrt{100-20}}{2}$$ $$Q=\frac{10\pm \sqrt{80}}{2}$$ $$Q=\frac{10\pm 4\sqrt{5}}{2}$$ $$Q=5\pm 2\sqrt{5}$$

Since output cannot be negative, we take the positive root: $$Q=5+2\sqrt{5}\text{(approximately 9.47)}$$

However, for simplicity, we can also check if the quadratic equation can be factored: $$Q^2-10Q+5=0$$ Using the quadratic formula again, we get: $$Q=5\pm 2\sqrt{5}$$ Since $Q$ must be positive, the equilibrium output is: $$Q=5+2\sqrt{5}\text{(or approximately 9.47)}$$

But for exactness, we keep it in terms of $\sqrt{5}$.

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2. Profit at Equilibrium Output

Profit ($\pi$) is given by: $$\pi =\text{TR}-\text{TC}$$ Substitute $Q=5+2\sqrt{5}$ into TR and TC:

First, compute $Q^2$: $$Q^2=(5+2\sqrt{5}{)}^2=25+20\sqrt{5}+20=45+20\sqrt{5}$$

Now, calculate TR: $$\text{TR}=120Q-5Q^2$$ $$=120(5+2\sqrt{5})-5(45+20\sqrt{5})$$ $$=600+240\sqrt{5}-225-100\sqrt{5}$$ $$=375+140\sqrt{5}$$

Now, calculate TC: $$\text{TC}=60+7Q^2$$ $$=60+7(45+20\sqrt{5})$$ $$=60+315+140\sqrt{5}$$ $$=375+140\sqrt{5}$$

Now, compute Profit ($\pi$): $$\pi =\text{TR}-\text{TC}$$ $$=(375+140\sqrt{5})-(375+140\sqrt{5})$$ $$=0$$

Conclusion: At equilibrium, profit is zero because TR = TC.

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Alternative Approach (Using Marginal Revenue and Marginal Cost)

For a more rigorous approach, we can find profit-maximizing output where Marginal Revenue (MR) = Marginal Cost (MC).

1. Marginal Revenue (MR)

$$\text{TR}=120Q-5Q^2$$ $$\text{MR}=\frac{d(\text{TR})}{dQ}=120-10Q$$

2. Marginal Cost (MC)

$$\text{TC}=60+7Q^2$$ $$\text{MC}=\frac{d(\text{TC})}{dQ}=14Q$$

Set MR = MC for profit maximization: $$120-10Q=14Q$$ $$120=24Q$$ $$Q=5$$

Now, check profit at $Q=5$: $$\text{TR}=120(5)-5(5{)}^2=600-125=475$$ $$\text{TC}=60+7(5{)}^2=60+175=235$$ $$\pi =\text{TR}-\text{TC}=475-235=240$$

But this contradicts the earlier result. The discrepancy arises because the first method assumed TR = TC (break-even), while the second method finds profit-maximizing output.

Clarification:

• The equilibrium output (where TR = TC) is $Q=5+2\sqrt{5}$ (≈9.47), yielding zero profit.
• The profit-maximizing output (where MR = MC) is $Q=5$, yielding maximum profit of 240.

Since the question asks for equilibrium level of output and profit, we use the first method (TR = TC), where profit is zero.

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Final Answer

• Equilibrium level of output: $Q=5+2\sqrt{5}$ (≈9.47 units)
• Profit at equilibrium: 0 (since TR = TC)