Respuesta :
Let x denote the length of the side of the garden which is covered fenced by a shed, and [tex] \frac{A}{x} [/tex] be the width of the garden.
The perimeter of a rectangle is given by 2(length + width)
i.e. [tex]2x + \frac{A}{x} = 84[/tex]
which gives:
[tex]A = 84x - 2x^2[/tex]
For the area to be maximum, the differentiation of A with respect to x must be equal to 0.
i.e. [tex] \frac{dA}{dx} =84-4x=0 \\ 4x=84 \\ x=21[/tex]
Therefore, the maximum area of the garden enclosed is given by
[tex]84(21)-2(21)^2=1764-2(441)=1764-882=882 \, yd^2[/tex]
The perimeter of a rectangle is given by 2(length + width)
i.e. [tex]2x + \frac{A}{x} = 84[/tex]
which gives:
[tex]A = 84x - 2x^2[/tex]
For the area to be maximum, the differentiation of A with respect to x must be equal to 0.
i.e. [tex] \frac{dA}{dx} =84-4x=0 \\ 4x=84 \\ x=21[/tex]
Therefore, the maximum area of the garden enclosed is given by
[tex]84(21)-2(21)^2=1764-2(441)=1764-882=882 \, yd^2[/tex]