I need some help with this question

Question

the total cost of producing a type of tractor is given by C(x)=21000−50x+0.05x^2, where x is the number of tractors produced. How many tractors should be produced to incur minimum cost?

Answer 1

Hello, thank you for taking the time to post your question!

Since the expression given is a quadratic/parabola that opens up, the minimum value is going to occur at the vertex. The vertex is at x = -b/2a, which for this expression would be

x = -(-50)/2(0.05) = 50/0.10 = 500

That means that the number of tractors that would incur the minimum cost is 500

I hope that helps get you moving in the right direction! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

C'(x)=0

50-.1x=0

.1x=50x

x=50/.1=500 tractors minimize cost

C(500)= 21000 -50(500)-+05(1000/2)^2= 21000-25000+12500=8,500

that's the differential calculus method

or

use algebra

complete the square, to find the vertex (minimum point) of the upward opening parabola

C(x)=.05x^2-50x+21000=0

factor out .05, then

add and subtract .05(1000/2)^2=.05(250,000)

.05(x^2 -1000x +250,000)+21,000 -(.05)250,000=0

-.05(x-500)^2 + 21000-12500

-.05(x-500)^2 +8,500 =0 in vertex form

a(x-h)^2 + k =0, where (h,k)=(x,C(x))

vertex = minimum point = (500,8,500)

h=500= cost minimizing output level=x=500 tractors

k =8,500 =C(500) = minimum cost