Calculus maxima and minima word question can't understand

In summary, the conversation discusses a word problem involving finding the minimum amount of fencing required to enclose and divide a rectangular field into two separate rectangular areas. The participants identify the variables and function involved, but struggle with the number of variables and how to solve the problem. Ultimately, one of the participants realizes their mistake and provides the correct function.
  • #1
singleton
121
0
Calculus maxima and minima word question...can't understand :(

A rectangular field is going to be enclosed and divided into two separate rectangular areas (not equal either). Find the minimum fencing required if the total area of the field is 1200m^2

(See the picture attached right now)

My answer so far:
Let x represent the width of the rectangular area's width in metres
Let y represent the length of the first rectangular area in metres
Let z represent the length of the second rectangular area in metres

I've identified that we want to minimize the total amount of fencing P
So the function is:
P = 3x + 2y + 2z

Given the total area = area of first rectangle + area of second rectangle
1200 = xy + xz

The only problem is that I have three variables and I think I'm supposed to somehow do this with two as we've only been taught how to use two so far. I could figure it out with two but I don't know what is wrong with my answer so far :( Have I wrongly interpreted the question?

thanks for any help! :cry:
 

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  • #2
I can't believe I did something so stupid! :(

I shouldn't even have y and z it should just be y

P = 3x + 2y

1200 = xy
x = 1200/y

P = 3(1200/y) + 2y
P = 3600/y + 2y

first derivative is -3600/y^2 + 2
second derivative is 7200/y^3
etc etc (I think)

blah I can't believe I did something so stupid :(
 
  • #3
singleton said:
blah I can't believe I did something so stupid :(

On the other hand, you appear to have the situation well in hand.
 

Related to Calculus maxima and minima word question can't understand

1. What is the definition of maximum and minimum in calculus?

In calculus, maximum and minimum refer to the highest and lowest values of a function, respectively. These points are also known as extrema, and they can be found by taking the derivative of the function and setting it equal to zero.

2. How do you find the local maxima and minima of a function?

To find the local maxima and minima of a function, you must first take the derivative of the function and set it equal to zero. Then, solve for the values of x that make the derivative equal to zero. Finally, plug these values into the original function to find the corresponding y-values, which are the local maxima and minima.

3. Can a function have more than one maximum or minimum?

Yes, a function can have multiple local maxima and minima. These points are also known as relative extrema. They occur when the derivative of the function changes from positive to negative or vice versa.

4. How do you know if a point is a maximum or minimum?

To determine if a point is a maximum or minimum, you can use the second derivative test. If the second derivative is positive, then the point is a minimum. If the second derivative is negative, then the point is a maximum. If the second derivative is zero, then the test is inconclusive.

5. How is calculus used to solve real-world problems involving maxima and minima?

Calculus is used to find the maximum or minimum values in real-world situations, such as finding the highest point on a roller coaster or the lowest cost for producing a certain number of products. By applying the principles of calculus, we can optimize these situations to find the most efficient or desirable outcome.

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