Answer:
Step-by-step explanation:
There's the two widths, x.
There's the four lengths, y.
4y+2x=600
Also, A=xy.
x=300-2y
A=y*(300-2y)
dA/dy=0= y*(-2)+(300-2y)
-2y-2y+300=0
y=300/4=150/2=75
x=300-2y=300-150=150
Then the long side is 150ft, the short side is 75ft.
Check to make sure that the sum of the lengths is 600: 300+300=600
And the maximum area is 150*75 ft2
Maximizing the area of a field, means we want to get the highest possible area of the field.
The dimensions that maximize the area are: 75ft by 150ft
The given parameters are:
[tex]A = 600[/tex] --- the area of the field
The area of the field is:
[tex]A =x \times y[/tex] -- area of rectangle
When the field is divided into 2 smaller plots, the area of the field becomes:
[tex]A = 2x + 4y[/tex]
Substitute 600 for A in [tex]A = 600[/tex]
[tex]2x + 4y = 600[/tex]
Divide through by 2
[tex]x + 2y = 300[/tex]
Make x the subject
[tex]x = 300 - 2y[/tex]
Substitute [tex]x = 300 - 2y[/tex] in [tex]A =x \times y[/tex]
[tex]A=(300 - 2y) \times y[/tex]
Open bracket
[tex]A=300y - 2y^2[/tex]
Differentiate with respect to y
[tex]A'=300 - 4y[/tex]
Set to 0
[tex]300 - 4y=0[/tex]
Collect like terms
[tex]4y = 300[/tex]
Divide both sides by 4
[tex]y = 75[/tex]
Recall that: [tex]x = 300 - 2y[/tex]
[tex]x =300 - 2 \times 75[/tex]
[tex]x = 150[/tex]
Hence, the dimensions that maximize the area are: 75ft by 150ft
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