1(a) a rectangular plot is bounded at the back by a river. No fence is needed along the river and there is to be an opening in front of about 24 meters. If the cost of fencing in front is $1500 per meter and the cost of fencing on the sides is $1000 per meter, find the dimensions of the largest plot which can be fenced for $300 000.
(b) a manufacturer wants to design an open box having a square base and a surface area of 108 cm^2. What dimension will produce a box with maximum volume?
Let each of the sides be x m
let the length along the front be y m
so fencing needed
= 2x + (y-24) , ----- you said we needed an opening of 24 m at the front
1500(y-24) + 1000(2x) = 300000
1500y - 36000 + 2000x = 300000
2000x + 1500y = 336000
20x + 15y = 3360
4x + 3y = 672
y = (672 - 4x)/3 = 224 - (4/3)x
Area = A = xy
= x(224 - (4/3)x) = 224x - (4/3)x^2
dA/dx = 224 - (8/3)x
= 0 for a max of A
(8/3)x = 224
8x = 672
x = 84
then y = 224 - (4/3)(84 = 112
The rectangle should have a frontal length of 112 m and 2 sides of 84 m each.
b) let the base be x by x
let the height be y
SA = x^2 + 4xy
x^2 + 4xy = 108
4xy = 108 - x^2
y = (108-x^2)/(4x)
volume = x^2 y
V = x^2(108 - x^2)/(4x) = 27x - (1/4)x^3
dV/dx = 27 - (3/4)x^2 = 0 for a max/min of V
(3/4)x^2 = 27
x^2 = 36
x = 6 , rejecting the negative answer
then y = (108-36)/24 = 3
max volume = x^2 y = 108 cm^3
-- which is not a coincidence, by the way.
Thanks ReinY
The second term in the expansion of (1-x)(1+2x)Λ4 is 19x. Find the value of n
(a) To find the dimensions of the largest plot that can be fenced for $300,000, we need to maximize the area of the rectangular plot while considering the cost of fencing. Let's denote the length of the plot as L and the width as W.
The cost of fencing the front is $1500 per meter, so the cost of fencing the front would be 1500 * 24 = $36,000.
The cost of fencing the sides is $1000 per meter, so the cost of fencing both sides would be 1000 * 2L = $2000L.
Given that the total cost of fencing is $300,000, we can write the equation:
300,000 = 36,000 + 2000L.
Simplifying the equation, we have:
2000L = 264,000.
Dividing both sides by 2000, we find:
L = 132.
Now, the width of the plot (W) can be found by considering that the total enclosed area should be maximized. The area of the enclosed region is L * W. Since there is no fencing required along the river, we can assume that the width (W) should be equal to the length of the front opening, which is 24 meters.
Therefore, the dimensions of the largest plot that can be fenced for $300,000 are L = 132 meters and W = 24 meters.
(b) To find the dimensions that will produce a box with the maximum volume, we need to maximize the volume while considering the surface area of the box. Let's denote the side length of the square base as s.
The surface area of the open box can be calculated by adding the area of the base, which is s^2, and the areas of the four sides, which are 4s * h.
Given that the surface area is 108 cm^2, we can write the equation:
108 = s^2 + 4s * h.
To find the dimensions that maximize the volume, we need to express the volume (V) in terms of a single variable. The volume of the box is given by V = s^2 * h.
We can isolate h from the surface area equation. Rearranging, we have:
108 - s^2 = 4s * h,
h = (108 - s^2) / (4s).
Now, we can substitute this expression for h into the volume equation:
V = s^2 * [(108 - s^2) / (4s)].
Simplifying, we get:
V = (1/4) * (108s - s^3).
To find the maximum volume, we can take the derivative of V with respect to s, set it equal to zero, and solve for s:
dV/ds = 0,
(1/4) * (108 - 3s^2) = 0,
108 - 3s^2 = 0,
3s^2 = 108,
s^2 = 36,
s = 6.
So, the side length of the square base that will produce a box with maximum volume is 6 cm.