# Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible.?

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Find the dimensions of a rectangle with perimeter 76 m whose area is as large as possible. (If both values are the same number, enter it into both blanks.)

m (smaller value)

m (larger value)

• Well,

this proof does not use derivatives

let a = 76/4 = 19m

then the square of side a=19 has the area

A = 19^2 = 381

let e >= 0 represent a variation

let the length of any rectangle of perimeter p = 76m be :

L = a + e/2

and

let the width of any rectangle of perimeter p = 76m be :

l = a – e/2

then we verify that the total perimeter is :

2L + 2l = 2(a + e/2) + 2(a – e/2)

= 2a + e + 2a – e

= 2a

= p = 76m

and the area of this rectangle is a function of e, so:

A(e) = (a+ e/2) * (a – e/2)

= a^2 – e^2/4

is less, whenever e > 0 (meaning e >=0 AND e =/=0 !!)

in other words, the area of any rectangle of perimeter p is less then the Maximum reached for the square of side :

a = p/4 = 19m

conclusion :

the largest area is reached for the square of side 76/4 = 19m

qed

hope it’ ll help !!

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