Bulk buying Brilliant shirts

We can solve this problem using derived and finding the minimum point of the function. if : C(n) = 0,04N² - 36N + 9000 d[C(n)]/dn = 0,08 - 36 To find critical points, let's equal this to zero. 0,08n - 36 = 0, n = 450 After testing the point, you 'll find out that 450 it's a minimum point.

Take the common 0.04 out of each of the terms and we get

C(N)=0.04(( $N^{2}$ - 900N + 225000)

take the 225000 out of the brackets and complete the square

C(N)=0.04( $N^{2}$ - 900N + 202500) + 9000 - 8100

C(N)=0.04( $N-450)^{2}$ + 900

therefore at N=450, the cost of the shirts will start 'going down', so it is the tipping point of the function.

It's a function. Now, to know the minimum value of Shirts we will need to use the $Xv$ = - $\frac{-b}{2*a}$ . Soon, replacing $Xv$ = - $\frac{-36}{2 * 0.04}$ = $\frac{36}{0.08}$ = $\boxed{450}$

This can be done via differential calculus. On first differentiation of the function C(N) it gives value of N. This value of N either gives maximum value of C(N) or minimum. So to find which it is, we need to have a double differentiation. If double differentiation value is positive that means it is the minimum value. So the value of N that we got in first differentiation is our answer.

The minimum point will be at the vertex of this upward facing parabola. The x coordinate of the vertex is at -b/2a which is 36/.08= 450

The easiest way to solve the given problem is to get the range of the function. Let h be the range.

h = -b / 2a

h = -(-36) / 2(0.04)

h = 36 / 0.08

h = 450

$\begin{aligned} C(N) & = 0.04N^2 - 36N + 9000 \\ C(N) & = 0.04(N^2 - 900N) + 9000 \\ & = 0.04(N^2 - 900N + 202500) + 900 \\ \Rightarrow C(N) & = 0.04(N-450)^2 + 900 \end{aligned}$

We note that $C(N)$ is always $>0$ and it is minimum of $900$ when $N=\boxed{450}$ .

Differentiate the equation C(N)=0.04N^{2 }- 36*N +9000 to get C'(N)=0.08N -36

At an inflection point the derivative of the function is zero.However, this may be either a maximum point or a minimum point.To find out, get the second derivative . C''(N)=0.08

The fact that it is positive shows that the point is a minimum point.

Therefore C'(N) =0.08N -36=0 and

N=450.

C(N) is a function of the variable N C(N) is minimum when the derivative of the function is = 0 Y' = 2 0.04 N -36 = 0 Thus N =450

[dC(N)/dN]= .08N-36=o ; Maxima and minima in calculus N=450

I think we all know that to mind the maximum or minimum value of f(x), we first equate f'(x) to zero. We then find the root of f'(x) = 0. Let x0 be the solution of the equation. We then find f"(x0). If this value is greater than 0, f(x) has a minimum at x0. Carrying out the above procedure for the above equation,

f(N) = 0.04N^{2} - 36N + 9000

f'(N) = 0.08N - 36 = 0, gives us N = 450.

f"(450) = 0.08 > 0. Hence f(N) has a minimum at N = 450.

As the function is C(N)=0.04N2-36N+9000 and to minimize the total cost, the value of the derivative of the function, C'(N) should be less than or equal to 0 i.e C'(N)= 2(0.04)N-36 < or = 0 = 0.08N - 36 = 0 N= 450

-b/2a