Find the value of x that minimizes y= 7x^2+\frac{1900}{x} for positive x.


Find the value of {eq}x {/eq} that minimizes {eq}y= 7x^2+\frac{1900}{x} {/eq} for positive {eq}x {/eq}.

Critical Points:

The critical point is to be found first for finding the point where the curve takes the maximum or the minimum values. If there is only a single point that came to be critical, then there is either of the two extremes and not the both.

Answer and Explanation:

In the problem, we have to find the value of {eq}x {/eq} that minimizes {eq}y= 7x^2+\frac{1900}{x} {/eq} for positive {eq}x {/eq}.

So here we will first find the critical points of the function using the differentiation a follows:

{eq}y'=0\\ \Rightarrow \frac{d}{dx}\left(7x^2+\frac{1900}{x}\right)=0\\ \Rightarrow 14x-\frac{1900}{x^2}=0\\ \Rightarrow 14x^3-1900=0\\ \Rightarrow x=\sqrt[3]{\frac{950}{7}}\\ {/eq}

So now at this point, the function is going to take the minimum value as this is the only critical point that we have got, so no need to do further tests.

Learn more about this topic:

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9

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