In this problem, $$ p $$ is the price per unit in dollars and $$ q $$ is the number of units. If the weekly demand function is $$ p=164-2 q^{2} $$ and the supply function before taxation is $$ p=20+6 q $$, what tax per item will maximize the total tax revenue? $$ \$ \square $$

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We begin with the given demand and supply functions. Without tax, the demand is

$$
p = 164 - 2q^2
$$

and the supply is

$$
p = 20 + 6q.
$$

When a tax of $$t$$ dollars per item is imposed on the suppliers, the suppliers’ net receipt becomes $$p - t$$. Thus their supply function becomes

$$
p - t = 20 + 6q \quad \Longrightarrow \quad p = 20 + 6q + t.
$$

At equilibrium the price paid by consumers from the demand side equals the price given by the supply side. Hence, we equate

$$
164 - 2q^2 = 20 + 6q + t.
$$

Rearrange the equation:

$$
164 - 2q^2 - 20 - 6q - t = 0,
$$

which simplifies to

$$
144 - 2q^2 - 6q - t = 0.
$$

Moving the terms involving $$q$$ and $$t$$ to one side yields

$$
2q^2 + 6q + t - 144 = 0.
$$

Solving for $$q$$ in terms of $$t$$, we write the quadratic equation in $$q$$:

$$
2q^2 + 6q + (t - 144) = 0.
$$

Since we seek the positive equilibrium $$q$$, we use the quadratic formula where the discriminant $$D$$ is

$$
D = 6^2 - 4\cdot2\cdot(t-144) = 36 - 8(t-144) = 36 - 8t + 1152 = 1188 - 8t.
$$

Thus, the positive equilibrium quantity is

$$
q = \frac{-6 + \sqrt{1188 - 8t}}{4}.
$$

The total tax revenue $$R$$ is the product of the tax per item $$t$$ and the quantity $$q$$:

$$
R(t) = t\cdot q = t\left(\frac{-6 + \sqrt{1188 - 8t}}{4}\right).
$$

To maximize the revenue $$R(t)$$, we differentiate with respect to $$t$$. Define

$$
R(t)=\frac{t}{4}\left(-6+\sqrt{1188-8t}\right).
$$

Differentiate using the product rule:

$$
R'(t)=\frac{1}{4}\left[\left(-6+\sqrt{1188-8t}\right) + t\left(\frac{d}{dt}\sqrt{1188-8t}\right)\right].
$$

Differentiate the square root term:

$$
\frac{d}{dt}\sqrt{1188-8t} = \frac{-8}{2\sqrt{1188-8t}} = -\frac{4}{\sqrt{1188-8t}}.
$$

Thus, the derivative becomes

$$
R'(t)=\frac{1}{4}\left[-6+\sqrt{1188-8t}-\frac{4t}{\sqrt{1188-8t}}\right].
$$

Set the derivative equal to zero to find critical points:

$$
-6+\sqrt{1188-8t}-\frac{4t}{\sqrt{1188-8t}} = 0.
$$

Let

$$
x=\sqrt{1188-8t}.
$$

Then the equation transforms to

$$
-6 + x - \frac{4t}{x} = 0 \quad \Longrightarrow \quad x - \frac{4t}{x} = 6.
$$

Multiplying through by $$x$$ yields

$$
x^2 - 4t = 6x.
$$

Recall that $$x^2 = 1188-8t$$, so substitute back:

$$
1188-8t - 4t = 6x \quad \Longrightarrow \quad 1188 - 12t = 6x.
$$

Divide both sides by 6:

$$
198 - 2t = x.
$$

But $$x = \sqrt{1188-8t}$$, so

$$
\sqrt{1188-8t} = 198 - 2t.
$$

Square both sides:

$$
1188-8t = (198-2t)^2.
$$

Compute the square:

$$
(198-2t)^2 = 198^2 - 2\cdot198\cdot2t + 4t^2 = 39204 - 792t + 4t^2.
$$

Thus,

$$
1188-8t = 39204 - 792t + 4t^2.
$$

Bring all terms to one side:

$$
4t^2 - 792t + 39204 + 8t - 1188 = 0,
$$

which simplifies to

$$
4t^2 - 784t + 38016 = 0.
$$

Divide the entire equation by 4:

$$
t^2 - 196t + 9504 = 0.
$$

Now use the quadratic formula:

$$
t = \frac{196 \pm \sqrt{196^2 - 4\cdot1\cdot9504}}{2}.
$$

Calculate the discriminant:

$$
196^2 - 4\cdot9504 = 38416 - 38016 = 400.
$$

Thus,

$$
t = \frac{196 \pm 20}{2}.
$$

There are two solutions:

$$
t = \frac{216}{2} = 108 \quad \text{or} \quad t = \frac{176}{2} = 88.
$$

To determine which value maximizes the total tax revenue, we check the equilibrium quantity and revenue for each case.

For $$t=88$$:

$$
q = \frac{-6 + \sqrt{1188-8\cdot88}}{4} = \frac{-6+\sqrt{1188-704}}{4} = \frac{-6+\sqrt{484}}{4} = \frac{-6+22}{4} = \frac{16}{4} = 4.
$$

Tax revenue:

$$
R = 88\cdot4 = 352.
$$

For $$t=108$$:

$$
q = \frac{-6 + \sqrt{1188-8\cdot108}}{4} = \frac{-6+\sqrt{1188-864}}{4} = \frac{-6+\sqrt{324}}{4} = \frac{-6+18}{4} = \frac{12}{4} = 3.
$$

Tax revenue:

$$
R = 108\cdot3 = 324.
$$

Since $$352 > 324$$, the tax per item that maximizes the total tax revenue is

$$
\boxed{88}.
$$
The tax per item that maximizes the total tax revenue is \$88.

In this problem, is the price per unit in dollars and is the number of units.
If the weekly demand function is and the supply function before taxation is , what tax per item will maximize the total tax revenue?

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In this problem, is the price per unit in dollars and is the number of units. If the weekly demand function is and the supply function before taxation is , what tax per item will maximize the total tax revenue?