Derivatives

[Frankyannehhh]

A rectangular lot has an area of 1600 sq. m. Find the least amount of fence that could be used to emclose the area

sol:
let x and t the dimensions of the lot
area : xy = 1600 sq.m
y = 1600/x
perimeter: 2(x+y)
= 2(x+1600/x)

now i used the derivative test
p'(x) = 2-(3200/x²) =0
= 3200 = 2x²
= x² = 1600
= 40

and now i'm stuck and dont know what to do. Ive searched across the internet and found out that the answer is 160m but im not sure if its correct

 

[Eduleo]

What you have done seems correct. You have found that x=40 for the perimeter.... what about the perimeter value? Is this perimeter a minimum or a maximum?

 

[Frankyannehhh]

What you have done seems correct. You have found that x=40 for the perimeter.... what about the perimeter value? Is this perimeter a minimum or a maximum?

so when i plugged the value of x
y = 2(40+ 1600/40)
y = 160 m
is that the answer?

 

[Eduleo]

Almost right... you have misstyped: "y" instead of "p" in the formula.... the fence size is the "p" (perimeter) not one of lot's sides.

 

[Deleted member 4993]

A rectangular lot has an area of 1600 sq. m. Find the least amount of fence that could be used to emclose the area

sol:
let x and t the dimensions of the lot
area : xy = 1600 sq.m
y = 1600/x
perimeter: 2(x+y)
= 2(x+1600/x)

now i used the derivative test
p'(x) = 2-(3200/x²) =0
= 3200 = 2x²
= x² = 1600
= 40

and now i'm stuck and dont know what to do. Ive searched across the internet and found out that the answer is 160m but im not sure if its correct

Thank you for showing your work in detail. You have calculated one side of the rectangle \(\displaystyle \ \to \ \) x = 40 m

Now calculate the other side of the rectangle 'y' the equation for 'y' in your work?

Then calculate the perimeter of the rectangle 'p'.

What is the equation for 'p' in your work?

Continue.....

 

[Frankyannehhh]

Almost right... you have misstyped: "y" instead of "p" in the formula.... the fence size is the "p" (perimeter) not one of lot's sides.

x and y are the sides
solved for x which is
x- 40m

y=1600/x
y= 1600/40
y= 400m

p=2(x+y)
p= 2(40+400)
p= 880m

am i right? lol hahaha

edit:
oh wait is 160m the correct answer but im wrong in labeling?

 

[HallsofIvy]

A rectangular lot has an area of 1600 sq. m. Find the least amount of fence that could be used to emclose the area

sol:
let x and t the dimensions of the lot
area : xy = 1600 sq.m
y = 1600/x
perimeter: 2(x+y)
= 2(x+1600/x)

now i used the derivative test
p'(x) = 2-(3200/x²) =0
= 3200 = 2x²
= x² = 1600
= 40

First, the way you wrote this is very bad- confusing to anyone reading and, eventually, could confuse yourself! You are fine through "p'(x)= x²= 1600" but the next line says "p'(x)= 40" which is incorrect. You mean "x= 40".

and now i'm stuck and dont know what to do. Ive searched across the internet and found out that the answer is 160m but im not sure if its correct

The answer to what question?? WHY did you set the derivative equal to 0? I have said that your last line should have been "x= 40". Perhaps you forgot that you were solving for x?! And what does "x" mean? What does it represent? You started by saying "let x and t the dimensions of the lot" (you meant "x and y" of course). That's good. A lot of people just write an equation without saying what the variables represent. So x is one of the dimensions of the lot (width or length- it doesn't matter which). You also have "y= 1600/x" so with x= 40, y= 1600/40= 40. Also remember the units. just saying "40" doesn't mean anything. Is it 40 feet? 40 inches? 40 miles? The problem told you that the are is "1600 m²" so the length and width are 40 meters.

Now, what did the question ask? What problem are you trying to solve? Don't forget answer the question asked!

"Find the least amount of fence that could be used to enclose the area." You have found that the lot has dimensions 40 m by 40 m so there are 4 sides, each 40 meters long. The length of the fence is 4(40)= 160 m.

 

[Frankyannehhh]

i

First, the way you wrote this is very bad- confusing to anyone reading and, eventually, could confuse yourself! You are fine through "p'(x)= x²= 1600" but the next line says "p'(x)= 40" which is incorrect. You mean "x= 40".


The answer to what question?? WHY did you set the derivative equal to 0? I have said that your last line should have been "x= 40". Perhaps you forgot that you were solving for x?! And what does "x" mean? What does it represent? You started by saying "let x and t the dimensions of the lot" (you meant "x and y" of course). That's good. A lot of people just write an equation without saying what the variables represent. So x is one of the dimensions of the lot (width or length- it doesn't matter which). You also have "y= 1600/x" so with x= 40, y= 1600/40= 40. Also remember the units. just saying "40" doesn't mean anything. Is it 40 feet? 40 inches? 40 miles? The problem told you that the are is "1600 m²" so the length and width are 40 meters.

Now, what did the question ask? What problem are you trying to solve? Don't forget answer the question asked!

"Find the least amount of fence that could be used to enclose the area." You have found that the lot has dimensions 40 m by 40 m so there are 4 sides, each 40 meters long. The length of the fence is 4(40)= 160 m.

THANK YOU, sorry for bothering it's my first time here, ill promise next time ill be good lol thanks