A company's revenue for selling x (thousand) items is given by R(x)=(35 x-x^2)/(x^2+35) (a) Find

step 1 I'm just going to jump into this problem where you can rewrite the function R of x 35x minus x s

A company's revenue for selling x (thousand) items is given...

A company's revenue for selling $x$ (thousand) items is given by $R(x)=\frac{35 x-x^{2}}{x^{2}+35}$ (a) Find the value of $x$ that maximizes the revenue and find the maximum revenue. (b) For any positive constant $c,$ find $x$ to maximize $R(x)=\frac{c x-x^{2}}{x^{2}+c}$

Key Concepts

Optimization in Calculus

Optimization in calculus involves using derivatives to determine the maximum or minimum values of a function. This technique is essential when aiming to maximize or minimize outcomes such as revenue by identifying the points at which the rate of change (the derivative) is zero, indicating potential extreme values.

Differentiation of Rational Functions

Differentiation of rational functions, which are ratios of two polynomials, typically involves the quotient rule. This rule provides a systematic way to find the derivative of a function expressed as the ratio of two differentiable functions, and it is crucial in solving optimization problems where the function represents quantities like revenue.

Critical Points

Critical points are values of the variable at which the derivative of a function is zero or undefined. In the context of revenue optimization, identifying these points is important because they represent candidates for local maximum or minimum values, which must be tested to determine which one yields the maximum revenue.

Parameter Optimization

Parameter optimization extends the concept of maximizing a function by incorporating constants or parameters in the function. This approach allows analysis of how changes in these parameters impact the optimal solution, which is useful in generalized models where the relationships involve variable constants impacting the outcome.

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