Suppose the revenue equation is of the form R = ax2 + bx + c where a, b and c are constants and x is the variable.
To have a maximum, either a must be negative or x must lie within fixed limits.
If a is negative then the maximum revenue is attained when x = -b/(2a). That is, find the value of R when x = -b/(2a).
If a is positive, then find the value of R when x is at each end point of its domain. One of them will be larger and that is the maximum value of the revenue.
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Yes, however not all quadratic equations can easily be solved by factoring, sometimes you can factor and sometimes it is easier to use the quadratic formula. Example: x2 + 4x + 4 This can be easily factored to (x + 2)(x +2) Therefore the answer is -2 by setting x +2 = 0 and solving for x This can be done using the quadratic equation and you would get the same results, however, it was much faster to factor instead.
( x - 2)( x - 6) = 0 x = 2 or x = 6
For any quadratic ax2 + bx + c = 0 we can find x by using the quadratic formulae: the quadratic formula is... [-b +- sqrt(b2 - 4(a)(c)) ] / 2a
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