Optimization: find zeros of a derivative

In summary: Thank you so much for your help!In summary, to find the maximum point of P(h)=-10h+4410-(6800/h), you need to set the derivative P'(h) equal to 0 and solve for h. This leads to h^2=680 and the maximum point is at h=82.5.
  • #1
BeccaHua
3
0

Homework Statement


Find the maximum point of P(h)=-10h+4410-(6800/h)


Homework Equations


P(h)=-10h+4410-(6800/h)

The Attempt at a Solution


P(h)=-10h+4410-(6800/h)
P'(h)=-10+(6800/h^2)
P'(0)=-10h^2+6800
10h^2=6800
Divide both sides by 10:
h^2=680
and sqrt both sides:
h=26.1
 
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  • #2
BeccaHua said:

Homework Statement


Find the maximum point of P(h)=-10h+4410-(6800/h)


Homework Equations


P(h)=-10h+4410-(6800/h)

The Attempt at a Solution


P(h)=-10h+4410-(6800/h)
P'(h)=-10+(6800/h^2)
P'(0)=-10h^2+6800
10h^2=6800
Divide both sides by 10:
h^2=680
and sqrt both sides:
h=26.1

h=sqrt(680) isn't the only root, there's also h=(-sqrt(680)). Or are you only considering h>0? You'll still want to consider what happens near h=0. What's your question?
 
  • #3
Oops sorry.. My question is how do you isolate h from P'(h)=-10+(6800/h^2)
I know the correct answer is h=82.5 but I'm not sure how to get there. I know that the sqrt of 6800 is 82.5 but then what happens with the -10?
 
  • #4
BeccaHua said:
Oops sorry.. My question is how do you isolate h from P'(h)=-10+(6800/h^2)
I know the correct answer is h=82.5 but I'm not sure how to get there. I know that the sqrt of 6800 is 82.5 but then what happens with the -10?

If you are trying to solve 0=-10+(6800/h^2) to find a point where the derivative equals 0 then I think you already did it correctly. That means 10=6800/h^2 so 10h^2=6800, h^2=680. The derivative isn't 0 at h=82.5. That's wrong.
 
  • #5
Okay thank you! The answer that was given was my teacher's answer and I was going crazy trying to figure out what I was doing wrong.
 

Related to Optimization: find zeros of a derivative

1. What is optimization?

Optimization is a mathematical process of finding the best possible solution for a problem, given a set of constraints. It involves maximizing or minimizing an objective function by adjusting certain variables.

2. What are zeros of a derivative?

Zeros of a derivative, also known as critical points, are the points on a graph where the derivative of a function is equal to zero. They represent the points where the slope of the function is either flat (horizontal) or changes direction (from increasing to decreasing or vice versa).

3. How do you find zeros of a derivative?

To find zeros of a derivative, you can set the derivative of the function equal to zero and solve for the variable. This will give you the x-values of the critical points. You can also use methods such as the first or second derivative test to determine whether these points are local maxima, local minima, or points of inflection.

4. Why is finding zeros of a derivative important?

Finding zeros of a derivative is important because it helps us identify important points on a graph, such as local maxima or minima, where the function has the highest or lowest values respectively. This information is useful in optimization problems, as it allows us to determine the best possible solution for a given objective function.

5. What are some real-world applications of finding zeros of a derivative?

Finding zeros of a derivative has many real-world applications, such as in economics, engineering, and statistics. For example, it can be used to find the optimal production level for a company to maximize profits, to determine the maximum weight a bridge can support, or to find the best fit line for a set of data points. Essentially, any problem that involves finding the best solution within a set of constraints can benefit from finding zeros of a derivative.

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