Question

The total cost C(g) of producing q goods is given by: C(q) = 0.01q^3- 0.6q^2 + 12qWhat is the maximum profit if each item is sold for 10 dollars? (Assume you sell everything you produce. Also note that you can only produce dwhole number of goods.)maximum profit = ___dollarsSuppose we fix production at 38 goods produced, and that they all sell when the price is 10 dollars each. Also suppose that for each 1 dollar increase in price, 1 fewer goods are sold (so if the price is 11 dollars, 37 of the 38 goods being produced are sold.)To maximize profit in this case, we should increase the price by ___dollars.(Enter zero if the price should remain at 10 dollars.)

Answer 1

$a)\ 241.68$

a) To maximize the profit, we'll need to find the Profit function. We know that each item is sold for $10 so the Revenue is R(q)= 10q the profit will be found by subtracting the cost from the Revenue:

$\begin{gathered} C(q)=0.01q^3-0.6q^2+12q \\ R(q)=qx,\Rightarrow R(10)=10q \\ P(q)=10q-(0.01q^3-0.6q^2+12q) \\ P(q)=-0.01q^3+0.6q^2-2q \end{gathered}$

Now, to maximize it we'll need to take the derivative of this Profit function and equate it to zero, applying the power rule like this :

$\begin{gathered} P(q)=-0.01q^3+0.6q^2-2q \\ P^{\prime}(q)=-0.03q^2+1.2q-2 \\ -0.03q^2+1.2q-2=0 \\ -0.03q^2\cdot \:100+1.2q\cdot \:100-2\cdot \:100=0\cdot \:100 \\ -3q^2+120q-200=0 \\ q_=\frac{-120\pm\sqrt{120^2-4\left(-3\right)\left(-200\right)}}{2\left(-3\right)} \\ q_1=\frac{10\left(6-\sqrt{30}\right)}{3}=1.74 \\ q_2=\frac{10\left(6+\sqrt{30}\right)}{3}=38.25 \end{gathered}$

Let's take the second derivative test to get to know which are we going to use:

$\begin{gathered} P^{\prime}^{\prime}(q)=-0.06q+1.2 \\ P"(1.74)=-0.06(1.74)+1.2=1.0956>0\text{ Minimum} \\ P"(38.25)=-0.06(38.25)+1.2=-1.0956<0,Maximum \end{gathered}$

So the Maximum profit is obtained if we sell it for 38 units (rounding off to the nearest whole), which yields:

$\begin{gathered} P(38)=-0.01(38)^3+0.6(38)^2-2(38) \\ P(38)=\ 241.68 \end{gathered}$