Respuesta :
Let's assume
length of rectangle is x
width of rectangle is y
now, we can draw rectangle
now, we can find fencing length
[tex] F=2x+3y [/tex]
we are given
total fencing length=400
so, we can set them equal
[tex] 400=2x+3y [/tex]
now, we can solve for y
[tex] 3y=400-2x [/tex]
[tex] y=\frac{400}{3} -\frac{2x}{3} [/tex]
now, we can find area
Area=length*width
[tex] A=xy [/tex]
we can plug back y
[tex] A=x(\frac{400}{3} -\frac{2x}{3}) [/tex]
[tex] A=\frac{400}{3}x -\frac{2x^2}{3} [/tex]
we have to maximise area
For maximising area , firstly we find derivative of area
and then we can set derivative =0 and then we solve for x
so, firstly we will find derivative
[tex] A'=\frac{400}{3}*1 -\frac{2*2x}{3} [/tex]
now, we can set it to 0
and then we solve for x
[tex] 0=\frac{400}{3}*1 -\frac{2*2x}{3} [/tex]
[tex] 400-4x=0 [/tex]
[tex] x=100 [/tex]
now, we can find y
[tex] y=\frac{400}{3} -\frac{2x}{3} [/tex]
[tex] y=\frac{400}{3} -\frac{2*100}{3} [/tex]
[tex] y=\frac{200}{3} [/tex]
so, dimensions are
length is 100 feet
width is [tex] \frac{200}{3} [/tex] feet.............Answer
Dimensions of a rectangle are
length = 100 ft
[tex]\rm width = \dfrac{200}{3}\;ft[/tex]
Step-by-step explanation:
Given :
Fencing length, F = 400 ft
Calculation :
Let the length of the rectangle be 'a' and width of the rectangle be 'b'.
Now Fencing length is,
F = 2a + 3b
400 = 2a + 3b
Now,
[tex]b = \dfrac{400}{3}-\dfrac{2a}{3}[/tex] ------ (1)
We know that,
[tex]\rm Area = length \times width[/tex]
[tex]\rm A = ab[/tex]
From equation (1),
[tex]\rm A = a (\dfrac{400-2a}{3})[/tex]
[tex]\rm A = \dfrac{400a}{3}-\dfrac{2a^2}{3}[/tex]
For maximizing enclosed are we have to find derivative of a
[tex]\rm A' = \dfrac{400}{3}-\dfrac{4a}{3}[/tex]
Now putting A' = 0, we get
a = 100 ft
From equation (1),
[tex]\rm b = \dfrac{200}{3}[/tex]
Therefore, dimensions of a rectangle are
length = 100 ft
[tex]\rm width = \dfrac{200}{3}\;ft[/tex]
For more information, refer the link given below
https://brainly.com/question/11897796?referrer=searchResults
