Please help solve this,
A farmer has 600m of fence and wants to enclose a rectangular field beside a river. Determine the dimensions of the fence field in which the maximum area is enclosed. (Fencing s required on only three sides: those that aren't next to the river.)
A=L*W
P=L*2W three sides
600m=L*2W
L=600-2W
substitute in the area
A=(600-2W)*W
A=600W-2W^2
convert the eq. to vertex form y=a(x-h)^2+k
A=-2W^2+600W
complete the square
A=-2(W^2-300W+22500)+22500*2
A=(W-150)^45000
w=150
substitute calculate....
L=600-2w
L=600-300 =300
dimension
150x300
Well, well, well, Mr. Farmer has a little fencing predicament, does he? Alright, let's put on our mathematical clown hats and figure this out!
So, we have a rectangular field with fencing required on only three sides, leaving the side next to the river all by its watery lonesome. Let's call the length of the field L and the width W.
To maximize the area, we need to find the maximum value of L x W. But we also have the constraint that the farmer has only 600 meters of fencing to work with. Oh boy, the drama!
Alright, let's see what we can do. Since there are three sides for the fencing, we have:
2L + W = 600
To make things easier, let's solve this equation for W:
W = 600 - 2L
Now, let's substitute this value of W in our area equation:
Area = L x W
= L x (600 - 2L)
= 600L - 2L^2
To find the maximum area, we need to find the value of L that makes the derivative of the area equation equal to zero. That's where the party gets interesting!
Taking the derivative of the area equation with respect to L, we get:
d(Area)/dL = 600 - 4L
Now, let's set it to zero and solve for L:
600 - 4L = 0
4L = 600
L = 150
Aha! We've found the length!
Now, let's plug this value of L back into our equation for W:
W = 600 - 2L
W = 600 - 2(150)
W = 300
Ta-dah! The dimensions of the field that will enclose the maximum area are 150 meters by 300 meters. Now, Mr. Farmer can juggle with joy!
A family wants to fence a rectangular play area alongside the wall of their house. The wall of their house bounds one side of the play area. If they want the play area to be exactly 400ft2, what is the least amount of fencing needed to make this? Round your answer to the nearest two decimal places.
To determine the dimensions of the rectangular field that will enclose the maximum area with only three sides fenced, we can use the concept of optimization.
Let's say the length of the rectangular field is L and the width is W. As given in the problem, three sides are fenced, which means that only two lengths and one width will be enclosed.
From the information provided, we know that the farmer has 600m of fence available. Since three sides need to be fenced, the total length of the three sides would be 600m.
Considering the three sides that are fenced, we have:
2L + W = 600
Now, we need to maximize the area of the rectangular field. The area of a rectangle is given by A = L * W.
To solve the problem, we can express one of the variables in terms of the other and substitute it into the area equation.
From the equation 2L + W = 600, we can rearrange it as follows:
W = 600 - 2L
Substituting this value of W into the area equation:
A = L * (600 - 2L)
A = 600L - 2L^2
To find the maximum area, we need to find the vertex of the quadratic equation. The x-coordinate of the vertex of a quadratic equation in the form of ax^2 + bx + c is given by x = -b / (2a).
For our equation A = 600L - 2L^2, the coefficient of L^2 is -2, and the coefficient of L is 600. So, the x-coordinate of the vertex is:
L = -600 / (2*(-2))
L = 150
Now that we have the value of L, we can substitute it back into the equation to find the corresponding width:
W = 600 - 2L
W = 600 - 2(150)
W = 300
Therefore, the dimensions of the fence field that will enclose the maximum area are:
Length (L) = 150m
Width (W) = 300m
A=L*W
P=L*2W three sides
600m=L*2W
L=600-2W
substitute in the area
A=(600-2W)*W
A=600W-2W^2
convert the eq. to vertex form y=a(x-h)^2+k
A=-2W^2+600W
complete the square
A=-2(W^2-300W+22500)+22500*2
A=(W-150)^45000
w=150
substitute calculate....
L=600-2w
L=600-300 =300
dimension
150x300