Respuesta :
Let:
x = Width of the fence that is perpendicular to the house
y = Width of the fence that is parallel to the house
1) Since y is parallel to the house, the width x needs to be counted twice and the width y only counts once. The total length of the perimeter is:
P = 2x + y
The perimeter is known to be 920 feet of fencing, thus:
2x + y = 920
2) From the equation above, we can solve for y as follows:
y = 920 - 2x
3) The area of a rectangle is the product of both dimensions:
[tex]A=x\cdot y[/tex]Substituting the expression for y:
[tex]\begin{gathered} A=x(920-2x) \\ A=920x-2x^2 \end{gathered}[/tex]To give the result of part 4, I need to solve part 5 first.
5) To create the maximum area, we take the first derivative and equate it to 0:
[tex]A^{\prime}=920-4x=0[/tex]Solving the equation:
x = 920/4
x = 230
The value of y is:
y = 920 - 2x
y = 920 - 460
y = 460
The dimensions are x = 230 feet and y = 460 feet
4)
Substituting into the formula of the area:
[tex]\begin{gathered} A=920\cdot230-2\cdot230^2 \\ \text{Operating:} \\ A=105,800 \end{gathered}[/tex]The maximum area is 105,800 square feet
