One side of a rectangle sheep pen is formed by a hedge. The other three sides are made using fencing. The length of the rectangle is x meters; 120 meters of fencing is available.
a) Show that the area of the rectangle is 1/2x(120-x)m^2. http://tinypic.com/r/2hpi0ir/5 <-this
b) Calculate the maximum possible area of the sheep pen.
i need to actually understand it. Questions on my test will be like this. please show work!
a) Show that the area of the rectangle is 1/2x(120-x)m^2. http://tinypic.com/r/2hpi0ir/5 <-this
b) Calculate the maximum possible area of the sheep pen.
i need to actually understand it. Questions on my test will be like this. please show work!
-
it's easy. You know that there is a rectangle which has four sides, one of which is x meters. the sum of the other three sides is 120. Since a rectangle has two short sides and two long sides we know that 120 - x = the width of the rectangle multiplied by 2. The formula of the area of the rectangle is length*width so xm* 1/2(120-x)m= 1/2x(120-x)m^2-
to solve the second point you need to calculate the derivative of the function you found
f(x)= 1/2x(120-x)=60x-x^2
f '(x)=60-2x
now you have to calculate f '(x)>0 --> 60-2x>0 -->x<30 Thos means that when f '(x) is positive f(x) is growing, where it's negative f(x) is decreasing. SInce the value X=30 is the last one in wich the function
f(x) is increasing before it starts to decrease, then it's the maximum of the function. so it's the value for wich we have the maximum area of the rectangle.
to solve the second point you need to calculate the derivative of the function you found
f(x)= 1/2x(120-x)=60x-x^2
f '(x)=60-2x
now you have to calculate f '(x)>0 --> 60-2x>0 -->x<30 Thos means that when f '(x) is positive f(x) is growing, where it's negative f(x) is decreasing. SInce the value X=30 is the last one in wich the function
f(x) is increasing before it starts to decrease, then it's the maximum of the function. so it's the value for wich we have the maximum area of the rectangle.
-
Questions of these type require you to use the given data to form one equation, which you will then substitute into another. Here we've been given the total fencing to be 120. Let the length of the hedge be x and its width be y
y
---
l l
l l x lol at diagram
l l
---
Total fencing = y+x+y (not including hedge) =120 (from question)
hence 2y+x=120<----IMPORTANT
We want a formula for area
A=xy (area of a rectangle), but if you notice, the actual answer is needed in terms of x only.
Because 2y+x=120, y=1/2(120-x), and we'll sub this into A
A=xy
A=x(1/2(120-x))
Thus A=1/2x(120-x)
To find max area just differentiate this and equate to zero and solve for x. Put this x value back into the A formula
y
---
l l
l l x lol at diagram
l l
---
Total fencing = y+x+y (not including hedge) =120 (from question)
hence 2y+x=120<----IMPORTANT
We want a formula for area
A=xy (area of a rectangle), but if you notice, the actual answer is needed in terms of x only.
Because 2y+x=120, y=1/2(120-x), and we'll sub this into A
A=xy
A=x(1/2(120-x))
Thus A=1/2x(120-x)
To find max area just differentiate this and equate to zero and solve for x. Put this x value back into the A formula