1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Svet_ta [14]
1 year ago
13

Help meeeeeeeee pleaseeeee rn rnnnn!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Amiraneli [1.4K]1 year ago
4 0

A rectangle that maximize the enclosed area has a length 800 yards and width is 800 yards. The maximum area is 640,000 square yards.

In the given question we have to find the maximum area of the rectangular area.

Liana has 3200 yards of fencing to enclose a rectangle.

So we know that the perimeter of rectangle is

P= 2(l+b)

Let length of the rectangle is x yards and width is y yards.

So the equation should be

2(x+y) = 3200

Divide by 2 on both side we get

x+y = 1600................(1)

As we know the area of rectangle is

A = xy

From the equation the value of y is 1600-x.

Now putting the value of y

A = x(1600-x)

A = 1600x-x^2

On differentiating

dA/dx = 1600-2x

Putting dA/dx=0

1600-2x=0

Subtract 1600 on both side we get

-2x= -1600

Divide by -2 on both side we get

x = 800 yards

Now putting the value of x in the y=1600-x

y=1600-800

y=800 yards

So the maximum area is

A= 800*800

A= 640,000 square yards

Hence, a rectangle that maximize the enclosed area has a length 800 yards and width is 800 yards. The maximum area is 640,000 square yards.

To learn more about perimeter and area of rectangle link is here

brainly.com/question/16572231

#SPJ1

You might be interested in
For the x-values 1,2,3, and so on, the y-values of a function form an arithmetic sequence that decreases in value. What type of
amid [387]

Answer:

Decreasing linear

Step-by-step explanation:

its c

8 0
2 years ago
What is the distance between points f (2,9) and G (2 , 6)​
solmaris [256]

use this formula

Step-by-step explanation:

Use the distance formula.

d = √[(x2 - x1)2 + (y2 - y1)2]

You have all of the coordinates. Plug in the numbers and evaluate.

6 0
4 years ago
Solve for x: 2x^5/3-22=464 a) x = 28 b) x = 27 c) x = ±27 d) x = ±28
Effectus [21]

Answer:

<h3>x = 27</h3>

Step-by-step explanation:

Given the expression {2x^{\frac{5}{3} }-22 = 464, we are to find the value of x in the expression. This is as shown below;

{2x^{\frac{5}{3} }-22 = 464

add 22 to both sides

{2x^{\frac{5}{3} }-22 = 464+22\\\\

{2x^{5/3} =486

divide both sides by 2

\frac{2x^{5/3}}{2} =\frac{486}{2}\\ x^{5/3} = 243

raise both sides to the power of 3/5

(x^{5/3})^{3/5} = (243)^{3/5}\\x = (\sqrt[5]{243}) ^3\\x = 3^3\\x = 27

Hence the value of x is 27

8 0
3 years ago
109⁰+118⁰+44⁰=× can you please help​
Paha777 [63]

Answer:

x = 271 degrees

Step-by-step explanation:

6 0
2 years ago
Jason earns $28 for every 4 lawns he mows.
aleksandr82 [10.1K]

Answer:

This could easily be answered with a culculator but the answer is 20.

Step-by-step explanation:

5x4=20

5 0
4 years ago
Other questions:
  • 1+4=5 2+5=12 3+6=21 8+11=
    8·2 answers
  • a shipping company charges x dollars to ship a package that weighs up to 1 pound and an additional fee based on the additional w
    5·1 answer
  • In a particular week a man receives 909.50 but 229.50 was for a overtime. If he works a basic week of 40 hours and overtime is p
    7·1 answer
  • Expression for the calculation double the product 4 and 7
    8·2 answers
  • What is 802.06 in expanded form
    12·2 answers
  • PROBABILITY QUESTION!!! PLEASE HELP!
    13·1 answer
  • Sam can do 120 jumping jacks in two minutes, how many jumping jacks can he complete in 5 mins.? (plz answer quick)
    14·2 answers
  • Place an operation sign between two of the digits to make the equation true 1 9 6 2 1 8=109
    14·2 answers
  • -2 + (-7) =<br> Help please
    10·1 answer
  • Could someone please help with this Alg 1 question ASAP?
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!