# Optimization variables problem

• physstudent1
In summary, the problem has to do with physics but it is from my calculus book, and for my calc class so I put it here. The homework statement is:A component is designed to slide a block of steel with weight W across a table and into a chute. The motion of the block is resisted by a frictional force proportional to its apparent weight. (Let k be the constant of proportionality.) Find the minimum force F needed to slide the block and find the corresponding value of theta. (Hint: FcosTheta is the force in the direction of the motion, and FsinTheta is the amount of force tending to lift the block. So the apparent weight is W-Fsint
physstudent1
This problem has to do with physics but it is from my calculus book, and for my calc class so I put it here:

## Homework Statement

"A component is designed to slide a block of steel with weight W across a table and into a chute. The motion of the block is resisted by a frictional force proportional to its apparent weight. (Let k be the constant of proportionality.) Find the minimum force F needed to slide the block and find the corresponding value of theta. (Hint: FcosTheta is the force in the direction of the motion, and FsinTheta is the amount of force tending to lift the block. So the apparent weight is W-Fsintheta.)"

## Homework Equations

Apparent weight = W-FsinTheta

## The Attempt at a Solution

I set the equation FcosTheta = k(w-Fsintheta)
I really don't know where to go from here. In every optimization problem we have found 2 equations a primary and a secondary and used the secondary to relate to get rid of variables. Can someone please point me in the right direction I'm pretty lost.

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Think back to basic calculus, sketching graphs of function, etc. If you have an equation y = f(x), how do you find the value(s) of x where the maximum or minimum value(s) of y occur?

Here you have an equation with two variables F and Theta, and you want the minimum value of F.

the answer is F = kW/(sqrt(k^2 +1)) if that helps anyone (its in the back of the book) I'm really not sure how to get there

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Do you know the answer to the question "how to find the min or max of a function y= f(x)" using calculus? If you drew a graph of a function y = f(x), what is the slope of the graph when y = a minimum or maximum? How do you find the slope of a graph?

When you do that, you will have another equation in F and Theta and you can solve the 2 equations.

yeah so I just take the derivative of my equation and set it equal to zero right? The only thing I'm not sure is what to set the equation equal to before I find the derivative should I make it F = (rest of equation).

I think my problem is in actually finding the derivative I get this for the derivative
F' = (-kw(-sinTheta+kcosTheta))/(cosTheta+ksinTheta)^2

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physstudent1 said:
yeah so I just take the derivative of my equation and set it equal to zero right? The only thing I'm not sure is what to set the equation equal to before I find the derivative should I make it F = (rest of equation).

Yes, take the derivative and set it equal to zero.

You can do it your way and write F = (rest of equation) if you want, but it's might be easier just to differentiate it as it is.

I'm going to write t for Theta 'cos I'm lazy so your equation was

F cos t = k(w - F sin t)
F cos t = kw - kF sin t
Differentiate with respect to t (remember k and w are constants)
Using the rule for differentiating a product of two functions f and g: d(f.g)/dt = f dg/dt + g df/dt

dF/dt cos t - F sin t = - k dF/ft sin t - kF cos t

At the minimun dF/dt = 0 so

- F sin t = - kF cos t
tan t = k

-kw(-sinTheta+kcosTheta)/(cosTheta+ksinTheta)^2 = 0

Multiply both sides by (cos Theta + k sin Theta)^2

-kw(-sin Theta + k cos Theta) = 0
k and w are not = 0 so -sin Theta + k cos Theta = 0
tan theta = k

I see! That makes a lot of sense I thought about deriving it that way at first, I just didn't realize to plug 0 in for dF/dt. Thanks alot! To get the Force I'm thinking I can plug in tan theta for k into the original equation, thanks for all the help.

physstudent1 said:
To get the Force I'm thinking I can plug in tan theta for k into the original equation, thanks for all the help.

You got it.

## What is an optimization variables problem?

An optimization variables problem is a type of mathematical problem that involves finding the best solution for a given set of variables, subject to constraints and limitations. This type of problem is common in many fields of science and engineering, where finding the optimal solution is important for achieving the desired outcome.

## What are the different types of optimization variables problems?

There are several types of optimization variables problems, including linear programming, nonlinear programming, integer programming, and dynamic programming. Each type of problem has its own set of techniques and algorithms for solving it, depending on the specific variables and constraints involved.

## How do you solve an optimization variables problem?

The process of solving an optimization variables problem involves several steps, including formulating the problem, selecting an appropriate optimization method, and using algorithms to find the best solution. The solution is typically reached through an iterative process of refining and adjusting the variables until the optimal solution is found.

## What are some applications of optimization variables problems?

Optimization variables problems have a wide range of applications in various fields, including economics, engineering, operations research, and computer science. They are commonly used for resource allocation, process optimization, scheduling, and decision-making in complex systems.

## What are the benefits of solving optimization variables problems?

Solving optimization variables problems can lead to significant benefits, such as improving efficiency, reducing costs, and increasing productivity. It can also help in making better decisions and optimizing processes to achieve desired outcomes. Additionally, the techniques used to solve these problems can be applied to a variety of real-world situations, making them a valuable skill for scientists and engineers.

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