For the cost function given by C(x)=3x2+4x+243C(x)=3x2+4x+243 a

Express the average cost function, ¯C(x)C¯(x). We do this by dividing the cost function, C(x)C(x), by x, or

¯C(x)=C(x)x¯C(x)=3x2+4x+243x¯C(x)=3x+4+243xC¯(x)=C(x)xC¯(x)=3x2+4x+243xC¯(x)=3x+4+243x

Now, we find the minimum of the average cost function. We first find x when the derivative of the average cost function is equal to zero, or ¯C′(x)=0C¯′(x)=0. We proceed with the solution.

¯C′(x)=0ddx(3x+4+243x)=03−243x2=03=243x2x2=81Take the positive root.x=9C¯′(x)=0ddx(3x+4+243x)=03−243x2=03=243x2x2=81Take the positive root.x=9

Therefore, the minimum average cost is at x = 9 or ¯C(9)C¯(9). We proceed with the solution.

¯C(9)=3(9)+4+2439=58C¯(9)=3(9)+4+2439=58

please see the attached file for the complete solution.