# BBandaRR's questions at Yahoo Answers regarding optimization

• MHB
• MarkFL
In summary: Other problems can be solved similarly, given the above theorem.In summary, in problem 5), the largest rectangle that can be inscribed in the ellipse 9x² + 16y² = 144 has a width of 4√2 and a height of 3√2. In problems 6) and 7), the maximum area of an isosceles trapezoid can be found by maximizing the square of the area function, with the width across the top being equal to (base + √(base² + 8slant²))/2.
MarkFL
Gold Member
MHB
Here are the questions:

CALCULUS worded problems HELP?

5) Find the dimensions of the largest rectangle that can be inscribed in the ellipse 9x² + 16y² = 144. The sides of the rectangle are parallel to the axes of the ellipse.

6) An isosceles trapezoid has a lower base of 16 cm and the sloping sides are each 8 cm. Find the width of the upper base for greatest area.

7) A trapezoidal gutter is to be made from a sheet of tin 22 cm wide by bending up the edges. If the base is 14 cm wide, what width across the top gives the greatest carrying capacity?

I have posted a link there to this topic so the OP can see my work.

Hello BBandaRR,

5.) I am assuming we wish to maximize the area of the resulting rectangle. By the symmetry of the ellipse and thus the inscribed rectangle, we need only concern ourselves with the first quadrant. So, let's first draw a diagram:

View attachment 1259

The area $A$ of the rectangle is our objective function:

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

subject to the constraint:

$$\displaystyle g(x,y)=9x^2+16y^2-144=0$$

Using a single-variable method, we may solve the constraint for either variable, and then substitute into the objective function to get the area as a function of one variable, and then use differentiation to find the maximum value. However, if we observe that maximizing the square of the area function times some constant is equivalent to maximizing the area itself, we may make our computations of the derivatives much simpler. So let us write:

$$\displaystyle \left(4A(x,y) \right)^2=f(x,y)=x^2\left(16y^2 \right)$$

By the constraint, we have:

$$\displaystyle 16y^2=144-9x^2$$

and so we may write:

$$\displaystyle f(x)=x^2\left(144-9x^2 \right)=144x^2-9x^4$$

Differentiating with respect to $x$, and equating the result to zero, we find:

$$\displaystyle f'(x)=288x-36x^3=36x\left(8-x^2 \right)=0$$

Since $x$ must be non-negative, we have the two critical values:

$$\displaystyle x=0,\,2\sqrt{2}$$

Computing the second derivative, we find:

$$\displaystyle f''(x)=288-108x^2$$

And so we know the critical value $x=0$ is associated with a minimum and the critical value $x=2\sqrt{2}$ is associated with a maximum.

Thus, for this value of $x$, we find:

$$\displaystyle y=\sqrt{\frac{144-9\cdot8}{16}}=\frac{3}{\sqrt{2}}$$

Now, using the fact that we wish to include all 4 quadrants in our final answer, this means we need to double $x$ and $y$ to find the dimensions of the rectangle of largest area that can be inscribed withing the given ellipse.

Hence, the width $w$ of the rectangle is:

$$\displaystyle w=2x=4\sqrt{2}$$

and the height $h$ of the rectangle is:

$$\displaystyle h=2y=\frac{6}{\sqrt{2}}=3\sqrt{2}$$

Our result would seem to suggest that the ratio of the width to the height of the rectangle is equal to the ratio of the length of the horizontal axis to the vertical axis of the ellipse in which the rectangle is inscribed.

Let's prove this conjecture using Lagrange multipliers. We have the objective function:

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

subject to the constraint:

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

Note: the constraint is derived from $$\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$.

And so, we obtain the following system:

$$\displaystyle y=\lambda\left(2b^2x \right)$$

$$\displaystyle x=\lambda\left(2a^2y \right)$$

This implies:

$$\displaystyle \lambda=\frac{y}{2b^2x}=\frac{x}{2a^2y}\implies \frac{x}{y}=\frac{a}{b}$$

Thus, our conjecture is true. Substitution into the constraint will yield the dimensions of the rectangle:

$$\displaystyle w=2x=\sqrt{2}a$$

$$\displaystyle h=2y=\sqrt{2}b$$

Problems 6.) and 7.) can both be solved by developing one theorem, that is to maximize the area of an isosceles trapezoid, given the base and the length of the slanted sides. Let's begin by drawing a diagram:

View attachment 1260

And so our objective function is:

$$\displaystyle A(h,x)=\frac{1}{2}(b+2x+b)h=(b+x)h$$

subject to the constraint:

$$\displaystyle g(h,x)=x^2+h^2-s^2=0$$

Solving the constraint for $h$, we find:

$$\displaystyle h=\sqrt{s^2-x^2}$$

And so we may write the objective function in one variable $x$:

$$\displaystyle A(x)=(b+x)\sqrt{s^2-x^2}$$

Now, again we may observe that maximizing the square of the area function will maximize the area function itself, and so we may write:

$$\displaystyle A^2(x)=f(x)=(b+x)^2\left(s^2-x^2 \right)$$

Differentiating with respect to $x$, and equating to zero, we find:

$$\displaystyle f'(x)=(b+x)^2(-2x)+\left(s^2-x^2 \right)2(b+x)=-2(b+x)\left(2x^2+bx-s^2 \right)=0$$

The only valid critical value (given that $0<x$) is obtained from the quadratic formula:

$$\displaystyle x=\frac{-b+\sqrt{b^2+8s^2}}{4}$$

The first derivative test shows that this critical value is associated with a maximum, if we observe that the factor $x+b$ must be positive, and the factors $-2\left(2x^2+bx-s^2 \right)$ are a parabola opening downward, and so is positive to the left of the larger root and negative to the right of this root.

Thus, the width $w$ across the top is:

$$\displaystyle w=b+2x=b+\frac{-b+\sqrt{b^2+8s^2}}{2}=\frac{b+\sqrt{b^2+8s^2}}{2}$$

So, for problem 6.) we have:

$$\displaystyle b=16\text{ cm},\,s=8\text{ cm}$$

Hence:

$$\displaystyle w=\frac{16+\sqrt{16^2+8\cdot8^2}}{2}\text{ cm}=8\left(1+\sqrt{3} \right)\text{ cm}$$

And for problem 7.) we have:

$$\displaystyle b=14\text{ cm},\,s=4\text{ cm}$$

Hence:

$$\displaystyle w=\frac{14+\sqrt{14^2+8\cdot4^2}}{2}\text{ cm}=16\text{ cm}$$

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THANKS MAN !

I want you to know how much I appreciate your detailed answers. It is rare to find such answerers like you. With your help I was able to get my assignment done, and be confident going to school. (Rofl)

I do wish to point out that we normally do not simply give complete solutions to problems posted here by our members. Our goal here for our registered members is actually loftier than that, in that we generally actively engage the person posting the question to take part in the process of solving the problem, so that the benefit is greater, i.e., more is learned.

I know I learn much more if someone guides me as I do something rather than just "pushes me aside" and instructs me to simply watch as they complete the task. It takes more time this way, but the benefit far outweighs this extra effort.

So, please feel free to post your questions here in the appropriate sub-forum, along with your work so we can see where you are stuck or may be going wrong. You will find we have many helpful and knowledgeable folks here who are glad to offer help. (Sun)

Hello BBandaRR,

Thank you for your questions regarding optimization. I am familiar with the concept of optimization and how it applies to mathematical problems.

For question 5, we can use the method of Lagrange multipliers to find the maximum area of the rectangle inscribed in the given ellipse. First, let's define the sides of the rectangle as x and y, with x being the length and y being the width. We can also define the ellipse equation as f(x,y) = 9x² + 16y² = 144. The constraint for the rectangle is that its sides must be parallel to the axes of the ellipse, so we can write this as g(x,y) = xy = constant.

Using the method of Lagrange multipliers, we can set up the following equation:
∇f(x,y) = λ∇g(x,y)
where ∇ is the gradient operator and λ is the Lagrange multiplier. This will give us a system of equations to solve for x, y, and λ.

Solving the system of equations, we get x = √(3/2) and y = √(2/3). Therefore, the dimensions of the largest rectangle that can be inscribed in the ellipse are √(3/2) by √(2/3).

For question 6, we can use the formula for the area of an isosceles trapezoid, which is A = (1/2)(a+b)(h), where a and b are the lengths of the bases and h is the height. In this case, we are given that the lower base is 16 cm and the sloping sides are each 8 cm. We can also define the width of the upper base as x.

Using the given information, we can write the following equation:
A = (1/2)(16+x)(8)
We can also write the constraint that the area must be maximized as g(x) = x = constant.

Using the method of Lagrange multipliers again, we can set up the following equation:
∇A(x) = λ∇g(x)
Solving this system of equations, we get x = 16 cm. Therefore, the width of the upper base for the greatest area is also 16 cm.

For question 7, we can use the formula for the area of a trapezoid, which is A =

## 1. What is optimization?

Optimization is the process of finding the best possible solution for a given problem or situation. It involves maximizing or minimizing a certain objective while adhering to certain constraints or limitations.

## 2. How is optimization used in real-life scenarios?

Optimization is used in a variety of real-life scenarios, such as in business to maximize profits, in logistics to minimize transportation costs, in engineering to design efficient structures, and in healthcare to optimize treatment plans.

## 3. What are some common optimization techniques?

Some common optimization techniques include linear programming, quadratic programming, dynamic programming, and evolutionary algorithms. These techniques use mathematical and computational methods to find the optimal solution.

## 4. What are the benefits of optimization?

The benefits of optimization include improved efficiency, cost savings, better decision-making, and increased competitiveness. By finding the best possible solution, optimization can help organizations achieve their goals more effectively.

## 5. Are there any limitations to optimization?

Yes, there are some limitations to optimization. It requires accurate data and assumptions, and it may not be able to consider all possible scenarios. Additionally, optimization may not always lead to the absolute best solution, but rather a good enough solution based on the given constraints.

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