i am lost completly i don't know what to do. please help me!!!!
a farmer is adding a rectangle corral to the side of a barn. the barn will form one side og the rectangle . the farmer has 16 m of fencing to use. conduct an investigation to determine the dimension of the corral with maxium area?
please help me!!
Did you make a sketch?
notice the farmer needs only one length, but two widths.
so 2w + l = 16
make a chart with three columns
one labeled "width", the next one "length" and the third "Area"
now fill it in starting with w=1, then w=2 etc
I see the following entries:
1 14 14
2 12 24
3 10 30
4 8 32 ....
can you see what is happening to the Area?
after the forth one the area is going backwards from before like 32,30,24?
does that mean 4 by 8 will give the gratest area?
yes
except you meant to say volume, not area, right?
isn;t it area becuase the third colmun is area
?
You all are dumb
25
Of course, I'm here to help! Let's break down the problem and conduct an investigation to determine the dimensions of the corral with the maximum area.
To start, we know that the corral is in the shape of a rectangle, and one side of the rectangle will be formed by the barn. The farmer has 16 meters of fencing to use to enclose the remaining three sides of the corral.
Let's assume the length of the corral is "L" meters and the width is "W" meters. Since one side of the rectangle is formed by the barn, its length will be equal to the width of the corral. Therefore, we can represent the dimensions of the corral as L x W.
Since the farmer has 16 meters of fencing available, we know that the total length of the three remaining sides is equal to 16 meters: L + W + L = 16.
To determine the dimensions of the corral that will maximize the area, we need to express the area of the rectangle in terms of L and W. The area of a rectangle is given by A = L x W.
In this case, we have two variables, L and W, and one equation, L + W + L = 16. We can rearrange the equation to express one variable in terms of the other: W = 16 - 2L.
Now, substitute W = 16 - 2L into the area equation: A = L x (16 - 2L) = 16L - 2L^2.
The area function is now expressed as a quadratic function: A = -2L^2 + 16L.
To find the maximum area, we need to determine the value of L that maximizes this quadratic function. One way to do this is by using calculus, specifically by finding the derivative of the area function and setting it equal to zero. However, since derivatives might be beyond the scope of this question, we can solve it using a simpler method.
We can create a table or plot points to compare areas for different values of L. Let's try some values:
When L = 0, the width W = 16 - 2(0) = 16, and the area A = 0 x W = 0.
When L = 4, the width W = 16 - 2(4) = 8, and the area A = 4 x 8 = 32.
When L = 8, the width W = 16 - 2(8) = 0, and the area A = 8 x 0 = 0.
From these values, we can see that at L = 4, the area is maximized with an area of 32 square meters.
Therefore, the dimensions of the corral with the maximum area would be 4 meters for the length and 8 meters for the width.
I hope this helps you understand how to approach this problem and find the solution. If you have any more questions, feel free to ask!