How to solve this out

A software company estimates that if it assigns x programmers to work on the project, it can develop a new product in T days, where
T = 100-30x+3x^2
How many programmers should the company assign in order to complete the development as quickly as possible ?

Explain how

T = 100-30x+3x^2 => rearrange
T = 3x^2 - 30x + 100 => this represents standard equation of a parabola in the form of:
y = ax^2 + bx + c, where a = 3 , b = -30 and c = 100
since a = 3 is positive the parabola opens upward (U shaped), that makes the vertex(turning point), the absolute(global) minima. hence you need to find the coordinates of the vertex:
x = -b/2a = [-(-30)] / (2 * 3) = 30/6 = 5 => number of programmers to do the job in a minimum time:
now you can calculate the T as well:
T = 3(5)^2 - 30(5) + 100
= 75 - 150 + 100 = 25 days => quickest time possible.
So your answer is x = 5 programmers.

You may also solve this using derivatives if you've already covered it:
T = 3x^2 - 30x + 100
dT/dx = 6x - 30 => set to zero
6x - 30 = 0
6x = 30
x = 30/6 = 5
T(5) = 25 => hence there is a critical point at(5 , 25), to find if it's Max or Min use the 2nd derivative test:
T" = 6 > 0 , therefore (5 , 25) is a global minima.

I hope this helps.

You are always welcome.

Regards.

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we have to minimize the value of T

here T = 100-30x+3x^2
dT/dx = -30 + 6x [ differentiating w.r.t x]....(i)
now put dT/dx = 0
so -30 + 6x = 0
or 6x = 30
or x =5

again differentiating equation (i) w.r.t. x

dT^2/d^2 x = 6 (+ve)

so at x = 5 this equation has a minima

so company should assign 5 programmers to complete work as soon as possible.

5. as quickly as possible, means T the smaller the better, so let it be 0. next, find the minimum point of the graph, 100-30x+3x^2. which is 5

It means T should be minimum.
100 - 30x +3x^2 = 100 + 3(x^2 - 10x) = 25 +3(x-5)^2 which is minimum if x = 5.
when x =5, T =25 days.

T ` (x) = - 30 + 6x = 0

x = 5

5 programmers required