Solution:
Given that the unit cost of manufacturing airplane engines in an aircraft factory is expressed by the function:
[tex]\begin{gathered} C(x)=1.2x^2-192x+20026 \\ where \\ C\Rightarrow cost\text{ in dollars} \\ x\Rightarrow number\text{ of engines made} \end{gathered}[/tex]To evaluate the number of engines to be made so as to minimize the unit cost,
step 1: Take the derivative of C(x) with respect to x.
[tex]C^{\prime}(x)=\frac{dC(x)}{dx}=2.4x-192[/tex]step 2: Evaluate the critical or stationary point of the C(x) function.
At, the stationary point, the derivative of C(x) equals zero.
Thus, at the critical point,
[tex]\begin{gathered} C^{\prime}(x)=0 \\ where \\ C^{\prime}(x)=2.4x-192 \\ \Rightarrow2.4x-192=0 \\ add\text{ 192 to both sides of the equation} \\ 2.4x-192+192=0+192 \\ \Rightarrow2.4x=192 \\ divide\text{ both sides by the coefficient of x, which is 2.4} \\ \frac{2.4x}{2.4}=\frac{192}{2.4} \\ \Rightarrow x=80 \end{gathered}[/tex]step 3: Take the second derivative of the C(x) function to determine the extreme points of the C(x) function.
[tex]C^{\prime}^{\prime}(x)=\frac{d^2C(x)}{dx}=2.4[/tex][tex]\begin{gathered} when\text{ C''\lparen x\rparen<0, we have a maximum point} \\ when\text{ C''\lparen x\rparen>0, we have a minimum point} \end{gathered}[/tex]Since the second derivative of C(x) is evaluated to be greater than zero, this implies that we have a minimum point or value of C(x).
Thus, for C(x) to be a minimum, we have the value of x to be 80.
Hence, number of engines to be made so as to minimize the unit cost is
[tex]Number\text{ of airplane engines}=80[/tex]