Answer:
Dimensions are length 15 feet and width 30 feet.
Maximum area of the plot is 450 feet².
Step-by-step explanation:
Let the length of the plot = x feet and width = y feet.
Since, the amount of fencing to be applied on the three sides of the plot is 60 feet.
We have,
[tex]2x + y = 60[/tex] i.e. [tex]y=60-2x[/tex]
Now, Area of the rectangular plot = Length × Width
i.e. Area of the rectangular plot = [tex]x \times y[/tex]
i.e. Area of the rectangular plot = [tex]x\times (60-2x)[/tex]
i.e. Area of the rectangular plot = [tex]-2x^2+60x[/tex]
Since, the maximum of the quadratic equation [tex]ax^2+bx+c[/tex] will be obtained at [tex]x= \frac{-b}{2a}[/tex].
Then, the maximum of the area of the plot is at the point [tex]x= \frac{-60}{2\times (-2)}[/tex] i.e. x= 15
Then, maximum area of the rectangular plot = [tex]-2x^2+60x[/tex]
i.e. maximum area of the rectangular plot = [tex]-2\times 15^2+60\times 15[/tex]
i.e. maximum area of the rectangular plot = -450+900 = 450 feet²
Hence, Maximum area of the plot is 450 feet².
Then,[tex]y=60-2\times 15[/tex] implies [tex]y=60-30[/tex] i.e. y= 30.
Hence, the maximum area is enclosed by the length 15 feet and width 30 feet.
