A rectangle has a perimeter of 32 in. Find the length and width of the rectangle under which the area is the largest.

a. Let the width to be x and the length to be y , then the quantity to be maximized is (expressed as a function of both x and y) A=_
b. The condition that x and y must satisfy is y=__

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boffeemadrid boffeemadrid · Answer 1 · 2020-11-10T18:00:03+01:00

Answer:

Length of the rectangle is 8 in and the breadth is 8 in.

a. [tex]A=xy[/tex]

b. [tex]y=16-x[/tex]

Step-by-step explanation:

Length = x

Breadth = y

Area is given by [tex]A=xy[/tex]

Perimeter of rectangle is given by

[tex]2(x+y)=32\\\Rightarrow x+y=16\\\Rightarrow y=16-x[/tex]

The condition that x and must satisfy is [tex]y=16-x[/tex]

So, area is

[tex]A(x)=x(16-x)\\\Rightarrow A(x)=16x-x^2[/tex]

Differentiating with respect to x we get

[tex]A'(x)=16-2x[/tex]

Equating with zero

[tex]0=16-2x\\\Rightarrow x=\dfrac{16}{2}\\\Rightarrow x=8[/tex]

Double derivative of A(x)

[tex]A''(x)=-2[/tex]

So [tex]A''(x)<0[/tex] which means [tex]A(x)[/tex] is maximum at [tex]x = 8[/tex]

[tex]y=16-x=16-8\\\Rightarrow y=8[/tex]

So length of the rectangle is 8 in and the breadth is 8 in.