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Marat540 [252]
3 years ago
11

Among all rectangles that have a perimeter of 182, find the dimensions of the one whose area is largest.

Mathematics
1 answer:
Evgen [1.6K]3 years ago
6 0

Answer:

<em>The largest rectangle of perimeter 182 is a square of side 45.5</em>

Step-by-step explanation:

<u>Maximization Using Derivatives</u>

The procedure consists in finding an appropriate function that depends on only one variable. Then, the first derivative of the function will be found, equated to 0 and find the maximum or minimum values.

Suppose we have a rectangle of dimensions x and y. The area of that rectangle is:

A=x.y

And the perimeter is

P=2x+2y

We know the perimeter is 182, thus

2x+2y=182

Simplifying

x+y=91

Solving for y

y=91-x

The area is

A=x.(91-x)=91x-x^2

Taking the derivative:

A'=91-2x

Equating to 0

91-2x=0

Solving

x=91/2=45.5

Finding y

y=91-x=45.5

The largest rectangle of perimeter 182 is a square of side 45.5

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the area of a rectangular field 4641 m squared. If the width of the field is 51 m, what is its length?
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❖ The length of the field is 91 m.

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~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

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