Calculus: Optimization Problem

In summary, the problem asks for the area of the largest rectangle that can be inscribed in the ellipse below. The quickest way to find this area is to use the pythagorean theorem and plug in the values for x and y. The problem asks for the area of the largest rectangle that can be inscribed in the ellipse below. The quickest way to find this area is to use the pythagorean theorem and plug in the values for x and y.
  • #1
RedBarchetta
50
1
Calculus: Optimization Problems

Homework Statement


Find the area of the largest rectangle that can be inscribed in the ellipse below.

I'm not quite sure where to start...first of all, how would you even enter this into a calculator to graph? On the TI-83, I only see one variable 'x' that you can enter...

I was reading a similar problem that said this was a lagrange multiplier problem. I was never taught this method. Is there any other way to do this?

________________________________________________________________________________________________________
2. The upper right-hand corner of a piece of paper, 14 in. by 10 in., as in the figure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose x to minimize y?
4-7-65.gif

Heres what I tried to do:

First I used the pythagorean theorem. a^2+b^2=y^2
We're trying to minimize y, so I set up the distance for each side.
(14-b)=top
(10-x)=side

Plug:
(14-b)^2+(10-x)^2=y^2

Since we can't solve for two variables, solve for one. Now what would I do? or would both variables be just "x"? like:
(14-x)=top
(10-x)=side

Thanks for the help!
 
Last edited by a moderator:
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  • #2
RedBarchetta said:

Homework Statement


Find the area of the largest rectangle that can be inscribed in the ellipse below.
http://www.webassign.net/www29/symImages/6/8/fbe8ac50ef95dfc0c03ab89315a6dc.gif

I'm not quite sure where to start...first of all, how would you even enter this into a calculator to graph? On the TI-83, I only see one variable 'x' that you can enter...

I was reading a similar problem that said this was a lagrange multiplier problem. I was never taught this method. Is there any other way to do this?


How abt assuming four points[tex] (a cos\theta , b sin \theta), (a cos\theta ,-b sin \theta),(-a cos\theta , b sin \theta),(-a cos\theta ,-b sin \theta) [/tex]

to be the vertices of that rectangle...then area =[tex]4absin\theta cos\theta=2absin 2\theta[/tex]

and then maximizing the area ? So 2ab ?
 
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  • #3
Your idea of starting a problem is to enter it into a calculator? Surely you should be able to graph an ellipse on paper and draw a rectangle in it?

The hard part would be using symmetry to show that the largest rectangle must have its sides parallel to the axes of the ellipse and that the rectangle is symmetric about both axes but you probably are expected to assume that. Assuming that, The coordinates of the vertices of the rectangle are (x0,y0), (-x0,y0), (-x0,-y0), and (x0,-y0) ( have subscripted to distinguish from the "variables" x and y). What is the area of such a rectangle, as a function of x0 and y0? Of course, x0 and y0 must also satisfy
[tex]\frac{x_0^2}{a^2}+ \frac{y_0^2}{b^2}= 1[/itex]
Yes, "Lagrange multiplier" would be the simplest way to solve that but you can also use the equation of the ellipse to reduce to one variable.
(Simple answer, by the way.)
 
  • #4
RedBarchetta said:
Plug:
(14-b)^2+(10-x)^2=y^2

This seems to be a pretty wrong equation to me. Which triangle are we concerned about here? (14-b) means, length-breadth. This can hardly mean any side of a triangle.

What you need to note for the 2nd problem is that, for a given 'x', only a particular value of 'y' is possible i.e. 'y' is a function of 'x' and 'x' alone. Try to arrive at 'y(x)'.
 
  • #5
Another approach:
My mathematics handbook gives the parametric form of an ellipse as:
x(t) = a cos t;
y(t) = b sin t;
The area of the inscribed rectangle is
area(t) = 4 x(t) y(t);
Solving the differential equation d/dt area(t)=0 gives the value of t at at the maximum
area. Using this value gives the area(t)=2 a b as shown in a previous reply.
 
  • #6
RedBarchetta said:
1. [URL]http://www.webassign.net/www29/symImages/6/8/fbe8ac50ef95dfc0c03ab89315a6dc.gif[/URL]2. The upper right-hand corner of a piece of paper, 14 in. by 10 in., as in the figure, is folded over to the bottom edge. How would you fold it so as to minimize the length of the fold? In other words, how would you choose x to minimize y?
[URL]http://www.webassign.net/scalcet/4-7-65.gif[/URL]
Heres what I tried to do:

First I used the pythagorean theorem. a^2+b^2=y^2
We're trying to minimize y, so I set up the distance for each side.
(14-b)=top
(10-x)=side

Plug:
(14-b)^2+(10-x)^2=y^2

Since we can't solve for two variables, solve for one. Now what would I do? or would both variables be just "x"? like:
(14-x)=top
(10-x)=side

Thanks for the help!
1.
Can't see 1st image (inaccessible) :(2.
Hint:
Hypotenuse in right-angled triangle - minimum - isoscles.
=>
Solution:
x=
10
y=
14.1421356
Explanation:

If done practically, the solution presents itself instantaneously.
http://goo.gl/Cirv3

Theoretically:
The triangle formed by the folded paper is a right-angled triangle (diag).
The minimum hypotenuse in a right-angled triangle is when both the sides are equal (isoscles).
so y is minimum when 10-x=0 i.e. when x=
10

so
y2 = x2 + x2
y2= 2 . x2
y= x . [tex]\sqrt[]{}[/tex]2
y = 10 (1.41421356)
y=14.14
 
Last edited by a moderator:
  • #7
Aruna is (respectfully) wrong.
The real minimum occurs when x=7.5, which gives y=12.99
This can be found legitimately (as opposed to aruna's hand-wavy argument) using similar triangles.
 

Related to Calculus: Optimization Problem

What is an optimization problem in calculus?

An optimization problem in calculus involves finding the maximum or minimum value of a function, subject to a set of constraints. It is used to solve real-world problems such as maximizing profit, minimizing cost, or finding the most efficient solution.

How is calculus used to solve optimization problems?

Calculus is used to solve optimization problems by finding the critical points of a function, which are points where the derivative is equal to zero. These points can then be evaluated to determine whether they correspond to a maximum or minimum value of the function.

What is the difference between a local and global maximum/minimum?

A local maximum or minimum is the highest or lowest value of a function in a specific interval, while a global maximum or minimum is the highest or lowest value of the function over its entire domain. In other words, a local maximum or minimum is a relative extremum, while a global maximum or minimum is an absolute extremum.

What are some common applications of optimization problems in different fields?

Optimization problems can be found in various fields, including economics, engineering, physics, and biology. Some common applications include finding the most efficient route for a delivery truck, determining the optimal design for a bridge, and maximizing the production of a chemical reaction.

What techniques are used to solve optimization problems in calculus?

There are several techniques used to solve optimization problems in calculus, including the first and second derivative tests, Lagrange multipliers, and the method of substitution. These techniques involve finding the critical points and evaluating them to determine the maximum or minimum value of the function.

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