Hello there. To solve this question, we'll have to remember some properties about functions.
Given the cost function C(q) on the production of q goods is given by the expression:
[tex]C(q)=0.01q^3-0.6q^2+12q[/tex]And suppose the price of selling is 10 dollars per unit.
The profit function is given by P(q), that is the difference between the amount sold and the cost, i.e.
[tex]P(q)=10q-0.01q^3+0.6q^2-12q=-0.01q^3+0.6q^2-2q[/tex]We want to maximize this function, considering that when we increase the price by 1 dollar, we sell less units, that is, from 38 units, we sell 37.
We start plugging q = 38 into it
[tex]P(38)=-0.01\cdot38^3+0.6\cdot38^2-2\cdot38[/tex]Then we introduce a new variable r that is the amount that should be increased/decreased (the sign will tell us) such that when q turns to q - r, the selling price goes from 10 to 10 + r, so that
[tex]P(q-r)=-0.01(q-r)^3+0.6(q-r)^2-2(q-r)[/tex]Plugging q as 38, call P(38 - r) = F(r) such that
[tex]F(r)=0.01r^3-0.54r^2-0.28r+241.68[/tex]Now we take its derivative with respect to r
[tex]F^{\prime}(r)=0.03r^2-1.08r-0.28[/tex]This might be zero, so that we find the critical points of F
[tex]\begin{gathered} F^{\prime}(r)=0.03r^2-1.08r-0.28=0 \\ \\ r=\frac{1.08\pm\sqrt{(-1.08)^2-4\cdot0.03\cdot(-0.28)}}{2\cdot0.03} \\ \\ r\approx-0.2574 \\ r\approx36.2574 \end{gathered}[/tex]And taking the second derivative of F, we get
[tex]F^{\prime}^{\prime}(r)=0.06r-1.08[/tex]Evaluating it at the roots of F'(r), we get
[tex]\begin{gathered} F^{\prime}^{\prime}(-0.2574)=0.06\cdot(-0.2574)-1.08<0 \\ F^{\prime}^{\prime}(36.2574)=0.06\cdot36.2574-1.08>0 \end{gathered}[/tex]So we maximize F(r) at r = -0.2574 or approximately -0.26
And we say that P(q - r) is maximized when the price is 10 + r = $9.74, giving $241.72 of profit.
The negative sign of r tell us that the price should decrease by 0.26 dollars.
