# Glamour's questions at Yahoo Answers regarding optimization with constraint

• MHB
• MarkFL
In summary, the two real numbers $x$ and $y$ are constrained to have the product $xy=200$. If we find that x and y exchange places without changing the function or the constraint, then an extremum will occur at $x=y$. However, if we look at the end-points of the domain, we find that x is at the global maximum at $x=10$ and y is at the global minimum at $y=10$.
MarkFL
Gold Member
MHB
Here are the questions:

I don't get this topic so can you guys explain these question as simply as possible? Thanks

View attachment 1900

I have posted a link there to this thread so the OP can view my work.

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Hello Glamour,

36.) We have a point $(x,y)$ which is constrained to line on the curve:

$$\displaystyle x^2-y^2=4$$

Our objective function, that is, what we are seeking to minimize, is the distance between the points $(x,y)$ and $(6,0)$. Now, we may simplify our calculations somewhat if we define our objective function to be the square of this distance. Minimizing the square of the distance will also minimize the distance.

So, our objective function is:

$$\displaystyle f(x,y)=(x-6)^2+y^2$$

Subject to the constraint:

$$\displaystyle g(x,y)=x^2-y^2-4=0$$

First, let's look at a single-variable method. If we solve the constraint for $y^2$, we find:

$$\displaystyle y^2=x^2-4$$

Now, substituting this into our objective function, we get a function of one variable $x$:

$$\displaystyle f(x)=(x-6)^2+x^2-4=x^2-12x+36+x^2-4=2x^2-12x+32=2\left(x^2-6x+16 \right)=2(x+2)(x-8)$$

Pre-Calculus method of optimization:

We observe that because the coefficient of the squared term is positive, this parabolic function opens upwards, thus the vertex will be the global minimum. The axis of symmetry must lie midway between the roots:

$$\displaystyle x=\frac{-2+8}{2}=3$$

Substituting for $x$ into the constraint, we find:

$$\displaystyle y^2=(3)^2-4=5$$

Thus:

$$\displaystyle y=\pm\sqrt{5}$$

And so we conclude the points on the given curve closest to the given point are:

$$\displaystyle \left(3,\pm\sqrt{5} \right)$$

Single variable calculus method of optimization:

Recall we have the objective function:

$$\displaystyle f(x)=2\left(x^2-6x+16 \right)$$

To find the critical value(s), we differentiate with respect to $x$ and equate the result to zero and solve for $x$:

$$\displaystyle f'(x)=2\left(2x-6 \right)=4(x-3)=0\implies x=3$$

Now, we see that the second derivative of $f$ is a positive constant, thus we may conclude the extrema associated with this critical value is the global minimum.

As before, we substitute this critical value into the constraint to obtain the critical points:

$$\displaystyle \left(3,\pm\sqrt{5} \right)$$

Multi-variable method of optimization (Lagrange Multipliers):

Recall our objective function is:

$$\displaystyle f(x,y)=(x-6)^2+y^2$$

Subject to the constraint:

$$\displaystyle g(x,y)=x^2-y^2-4=0$$

This gives rise to the system:

$$\displaystyle 2(x-6)=\lambda(2x)$$

$$\displaystyle 2y=\lambda(-2y)$$

The second equation implies $\lambda=-1$, and so the first equation becomes:

$$\displaystyle x-6=-x\implies x=3$$

And as before, we substitute this critical value into the constraint to obtain the critical points:

$$\displaystyle \left(3,\pm\sqrt{5} \right)$$

37.) Let's let $x$ and $y$ be the two positive real numbers. Our objective function is their product:

$$\displaystyle f(x,y)=xy$$

And they are subject to the constraint:

$$\displaystyle x^2+y^2=200$$

Now, if we notice that the two variables possesses cyclic symmetry, that is we may exchange them without changing either the objective function or the constraint, then we may conclude immediately that an extremum will occur when $x=y$. Substituting for $y$ into the constraint, we find:

$$\displaystyle x^2+x^2=200$$

$$\displaystyle x^2=100$$

And taking the positive root, we conclude:

$$\displaystyle x=y=10$$

But, if we observe that as either $x$ or $y$ approaches $0$, the other variable must approach $10\sqrt{2}$ and the objective function will then approach zero in either case.

And so we may write:

$$\displaystyle f_{\max}=f(10,10)=10^2=100$$

$$\displaystyle f_{\min}=\lim_{x\to0}(xy)=0$$

However, let's explore how to work this without appealing to cyclic symmetry.

If we solve the constraint for $y$, taking the positive root, we find:

$$\displaystyle y=\sqrt{200-x^2}$$

Substituting this into our objective function, we find:

$$\displaystyle f(x)=x\sqrt{200-x^2}$$

Differentiating with respect to $x$ (using the Product Rule) and equating the result to zero, we find:

$$\displaystyle f'(x)=x\left(\frac{-2x}{2\sqrt{200-x^2}} \right)+(1)\sqrt{200-x^2}=\frac{2\left(100-x^2 \right)}{\sqrt{200-x^2}}=0$$

Now, we observe that the positive root of the denominator is an end-points of the objective function's domain, which is $0<x\le10\sqrt{2}$ and the (positive) root of the numerator is within the domain, so we have 3 critical values to check:

$$\displaystyle f(0)=0$$

$$\displaystyle f(10)=100$$

$$\displaystyle f\left(10\sqrt{2} \right)=0$$

Based on this, we may conclude:

$$\displaystyle f_{\max}=100$$

$$\displaystyle f_{\min}=0$$

Now, let's look at Lagrange multipliers. We have the objective function:

$$\displaystyle f(x,y)=xy$$

Subject to the constraint:

$$\displaystyle g(x,y)=x^2+y^2-200=0$$

Thus, we obtain the system:

$$\displaystyle y=\lambda(2x)$$

$$\displaystyle x=\lambda(2y)$$

This implies:

$$\displaystyle y^2=x^2$$

Substituting this into the constraint, we find:

$$\displaystyle x^2=100$$

And taking the positive root, we obtain

$$\displaystyle x=y=10$$

Now, as in the other methods we explored, we need to look at the end-points of the domain, to find the above critical point $(x,y)=(10,10)$ is at the global maximum, and the end-point values give the global minimum. As so, as before, we then find:

$$\displaystyle f_{\max}=100$$

$$\displaystyle f_{\min}=0$$

## 1. What is optimization with constraint?

Optimization with constraint is a problem-solving technique used in various fields such as engineering and mathematics. It involves finding the best possible solution to a problem while considering certain limitations or constraints.

## 2. How is optimization with constraint different from traditional optimization?

Traditional optimization does not take into account any constraints, meaning it focuses solely on finding the optimal solution without regard to limitations. On the other hand, optimization with constraint considers constraints as an integral part of the problem and aims to find the best solution that satisfies these constraints.

## 3. What are the common types of constraints in optimization?

The most common types of constraints in optimization include physical limitations such as resource availability or budget constraints, logical constraints such as time restrictions, and functional constraints such as design specifications or performance requirements.

## 4. How do you approach optimization with constraint?

To approach optimization with constraint, you must first clearly define the problem and identify the constraints. Next, you will need to formulate a mathematical model that represents the problem and its constraints. Finally, you can use various optimization techniques such as linear programming or genetic algorithms to find the best solution.

## 5. What are the benefits of using optimization with constraint?

Optimization with constraint allows for more realistic and practical solutions to problems by considering real-world limitations. It also helps in decision-making by providing the best possible solution within the given constraints. Additionally, it can save time and resources by identifying potential issues or conflicts early on in the problem-solving process.

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