On the y-axis, normally.
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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.
If x is the unknown or variable in an equation it can have many possible maximum or minimum values
Take the derivative of the function.By plugging a value into the derivative, you can find the instantaneous velocity.By setting the derivative equal to zero and solving, you can find the maximums and/or minimums.Example:Find the instantaneous velocity at x = 3 and find the maximum height.f(x) = -x2 + 4f'(x) = -2xf'(3) = -2*3 = -6So the instantaneous velocity is -6.0 = -2x0 = xSo the maximum height occurs at x = 0f(0) = -02 + 4 = 4So the maximum height is 4.
The only variable on the right hand side is sin(x). The maximum value of sin(x) is 1. So, the max value of 3sin(x) is 3*1 = 3 and so, the max value of 3sin(x) + 2 is 3+2 = 5.
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