The function c(x) = 600x – 0.3x2 represents total costs for a company to produce a product, where c is the total cost in dollars and x is the number of units sold.

C(x)=600x-0.3x²
1)we calculate the first derivative
C´(x)=600-0.6x

2)We math the firts derivative to 0; and we find out the value of "x"
600-0.6x=0
-0.6x=-600
x=-600/-0.6
x=1000

3) we calculate the second derivative:
C´´(x)=-0.6;  because -0.6>0; we have a maximun at x=1000.

4)We find out the maximum cost.

C(1000)=600(1000)-0.3(1000)²
C(1000)=600000-300000
C(1000)=300000

the maximum cost is $300000

C(x) = 600x - 0.3x²

At maximum cost differential of C(x) with respect to x = 0

dC/dx = 600*1x¹ ⁻ ¹ -  2*0.3x² ⁻ ¹ = 0

600*1x¹ ⁻ ¹ -  2*0.3x² ⁻ ¹ = 0

600x⁰ -  0.6x¹ = 0,   x⁰ = 1

600 -  0.6x = 0

600 = 0.6x

x = 600/0.6

x = 1000

The maximum cost occurs at x = 1000.

To actually confirm it is a maximum, the second differential must be < 0

dC/dx = 600 -  0.6x

d²C/dx² = 0 - 1*0.6x¹ ⁻ ¹ = -0.6x⁰ = -0.6*1 = -0.6

d²C/dx² = -0.6

The second differential -0.6 < 0, so it is a maximum point.

But the value of C(x) at the value of x = 1000

C(x) = 600x - 0.3x²

C(1000) = 600*1000 - 0.3*1000² = 300 000

The maximum cost is 300 000 units, at a quantity of 1000 units.