While searching for the minimum of
ƒ(x) � [x2 � (x � 1)2][x2 � (x � 1)2] 1 2 1 2
we terminate at the following points:
(a) x(1) � [0, 0]T
(b) x(2) � [0, 1]T
(c) x(3) � [0, �1]T
(d) x(4) � [1, 1]T
Classify each point.
AND
3.17. Suppose you are a fill-dirt contractor and you have been hired to transport
400 yd3 of material across a large river. To transport the material,
a container must be constructed. The following cost data are available:
Each round trip costs $4.20 and material costs for the container are
Bottom 20.00/yd2
Sides 5.00/yd2
Ends 20.00/yd2
Design a container to minimize cost that will perform the assigned task.
sry
Suppose you are a fill-dirt contractor and you have been hired to transport
400 yd3 of material across a large river. To transport the material,
a container must be constructed. The following cost data are available:
Each round trip costs $4.20 and material costs for the container are
Bottom 20.00/yd2
Sides 5.00/yd2
Ends 20.00/yd2
Design a container to minimize cost that will perform the assigned task.
To classify each point in the first question, we need to determine the nature of each point according to the function's properties. We can do this by evaluating the second derivative test at each point.
(a) x(1) = [0, 0]T:
To classify this point, we need to find the second derivative of ƒ(x). After finding the second derivative, substitute the values of x(1) into it. If the result is positive, x(1) is a local minimum. If the result is negative, x(1) is a local maximum. If the result is zero, the test is inconclusive.
(b) x(2) = [0, 1]T:
To classify this point, again find the second derivative of ƒ(x) and substitute the values of x(2) into it.
(c) x(3) = [0, -1]T:
Similar to the above steps, find the second derivative of ƒ(x) and substitute the values of x(3) into it.
(d) x(4) = [1, 1]T:
Once again, find the second derivative of ƒ(x) and substitute the values of x(4) into it.
By following these steps, you can classify each point based on the nature of the function.
Regarding the second question, to design a container to minimize cost, we need to determine the dimensions that will minimize the cost function. Let's break down the cost into separate components:
1. Cost of the bottom: This cost is given by the area of the bottom multiplied by the cost per square yard.
2. Cost of the sides: This cost is given by the sum of the areas of the sides multiplied by the cost per square yard.
3. Cost of the ends: This cost is given by the sum of the areas of the ends multiplied by the cost per square yard.
4. Round trip transportation cost: This cost is fixed at $4.20 per round trip, regardless of the dimensions of the container.
To minimize the cost, we need to find the dimensions (length, width, and height) that minimize the sum of these costs. This can be achieved by setting up an optimization problem with the constraints that the volume of the container should be 400 yd³.
Let's denote the dimensions as l, w, and h. Then the volume constraint can be expressed as l * w * h = 400.
Next, we need to express the cost as a function of these variables. The total cost would be the sum of the cost of the bottom, sides, ends, and the transportation cost. We can write this as:
Cost = Cost of bottom + Cost of sides + Cost of ends + Transportation cost
To find the minimum, you can use calculus techniques such as partial derivatives or Lagrange multipliers to optimize the cost function subject to the volume constraint.
By solving this optimization problem, you can find the dimensions of the container that will minimize the cost while meeting the volume requirement.