# Calculus Max. and Min. Question

• laker_gurl3
In summary, the lifeguard has 400m of rope to create a rectangular restricted swimming area on a public beach. To maximize the swimming area, the dimensions of the rectangle should be 100m by 200m. However, to ensure the safety of the swimmers, the lifeguard decides that nobody should be more than 50m from shore. In this case, the dimensions of the swimming area should be 50m by 300m. To find these dimensions, the area function of the rectangle, which is expressed in terms of one side, is maximized using the derivative. This gives the value for the side, which can then be used to find the other side and thus the dimensions of the swimming area.
laker_gurl3
The lifeguard at a public beach has 400m of rope available to lay out a rectangular restricted swimming area using the straight shoreline as one side of the triangle.
a.) If she wants to maximize the swimming area, what will the dimensions of the rectangle be?
b.)To ensure the safety of the swimmers, she decides that nobody should be more than 50m from shore. What should the dimensions of the swimming area be with this added restriction?...

What do i doo?

A is supposed to be 100m by 200m..and b is 50m by 300m...i can't get those at all though!

First, repair the statement of your problem. The area is in the shape of a rectangle, not a triangle

a) Three sides of the rectangle are formed by the rope. Let x = the length of the side parallel to the shore. Figure out how to express the other two sides in terms of x. Express the area of the rectangle in terms of x. Maximize that area. You will need the derivative of the area set equal to zero.

b) This is a simple calculation based on your expression of the area in terms of x. What is the value of x in this case?

Last edited:
oops yah ur right, the last word of the problem should have been rectangle..

The rectangle has sides a,a,b,b.Okay?From the text of the problem u're given that

$$2a+b=400$$ and you're asked about the values of "a" and "b" to maximize the area,i.e.the product $a\cdot b$.

From the constraint,u find that $b=400-2a$ and so the area function which needs to be maximized becomes

$$S=a(400-2a)$$

Find "a" for which the function S is maximum...

Daniel.

## What is the definition of a maximum and minimum point in calculus?

In calculus, a maximum point is the highest value of a function at a specific point, while a minimum point is the lowest value of a function at a specific point.

## How are maximum and minimum points identified in calculus?

To identify maximum and minimum points in calculus, we use the first and second derivative tests. The first derivative test involves finding the critical points of a function, and then testing the sign of the derivative to determine if it is a maximum or minimum point. The second derivative test involves taking the second derivative of the function at the critical point and determining its concavity to identify if it is a maximum or minimum point.

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

A local maximum or minimum point is the highest or lowest value of a function in a specific interval, while a global maximum or minimum point is the highest or lowest value of a function in its entire domain. In other words, a local maximum or minimum point is a point where the function is at its highest or lowest value within a small region, while a global maximum or minimum point is the overall highest or lowest value of the entire function.

## How do you graph a function with maximum and minimum points?

To graph a function with maximum and minimum points, we first identify the critical points and determine if they are maximum or minimum points using the first and second derivative tests. Then, we plot these points on the graph and connect them to create a curve. Depending on the function, the curve may have a maximum or minimum point at the end or at a turning point.

## What real-life applications use maximum and minimum points in calculus?

Maximum and minimum points in calculus are used in various real-life applications, such as optimization problems in engineering, economics, and physics. For example, a company may use calculus to determine the maximum profit they can make based on different production levels. In physics, maximum and minimum points are used to determine the maximum height or speed of an object in motion.

• General Math
Replies
1
Views
1K
• Calculus
Replies
8
Views
3K