Question 164164
N(x)=-0.4x^2+8x+12


we can use differentiation for this problem to find the optimum value
the optimum value is reached when 
{{{dN(x)/dx = 0}}}
-0.4*2*x + 8 = 0
<=> -0.8x + 8 = 0
<=> -0.8x = -8
<=> x = 8/0.8 = 10
because the coefficient of x^2 is negative, the function will have a maximum value.
the maximum value is:
{{{N(10) = -0.4*10^2 + 8*10 + 12}}}
N(10) = -0.4*100 + 80 + 12
= -40 + 80 + 12
= 52



or in another way, to find the maximum or minimum value of a quadratic function, this formula can be used:
because the coefficient of x^2 is negative, the function will have a maximum value.
the maximum value = {{{D/(-4a)}}} -> {{{D = b^2 - 4*a*c}}}


from the function above, the value of a = -0.4, b = 8, c = 12
so the maximum value
= {{{(8^2 - 4*(-0.4)*12)/(-4*(-0.4))}}}
= {{{(64 + 19.2)/1.6}}}
= 83.2/1.6
= 52