Area=legnth times width
area=3/5
legnth=7/8
width=x
3/5=7/8 times x
divide both sides by 7/8 (aka multiply both sides by 8/7 to cancel fraction because 7/8 times 8/7=56/56=1)
8/7 times 3/5=56/56 times x
24/35=1 times x
24/35=x
width=24/35 mile
Answer & Explanation Included:
f(x)= 2x - 4
f(5)= 2(5) - 4
f(5)= 10 - 4
f(5)= 6
Answer:
B) 0.6%
Step-by-step explanation:
9/1500 = 0.006 or 0.6%
What makes a band a band?
Here’s the hard part. We always want the problem structured in a particular way. Here, we are choosing to maximize f (x, y) by choice of x and y .
The function g(x,y) represents a restriction or series of restrictions on our possible actions.
The setup for this problem is written as l(x,y)= f(x,y)+λg(x,y)
For example, a common economic problem is the consumer choice decision. Households are selecting consumption of various goods. However, consumers are not allowed to spend more than their income (otherwise they would buy infinite amounts of everything!!). Let’s set up the consumer’s problem:
Suppose that consumers are choosing between Apples (A) and Bananas (B). We have a utility function that describes levels of utility for every combination of Apples and Bananas.
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A 2 B 2 = Well being from consuming (A) Apples and (B) Bananas.
Next we need a set pf prices. Suppose that Apples cost $4 apiece and Bananas cost $2 apiece. Further, assume that this consumer has $120 available to spend. They the income constraint is
$2B+$4A≤$120
However, they problem requires that the constraint be in the form g(x, y)≥ 0. In
the above expression, subtract $2B and $4A from both sides. Now we have 0≤$120−$2B−$4A
g(A, B) Now, we can write out the lagrangian
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l(A,B)= A2 B2 +λ(120−2B−4A)
f (A, B) g(A, B)
Step II: Take the partial derivative with respect to each variable
We have a function of two variables that we wish to maximize. Therefore, there will be two first order conditions (two partial derivatives that are set equal to zero).
In this case, our function is
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l(A,B)= A2 B2 +λ(120−2B−4A)
Take the derivative with respect to A (treating B as a constant) and then take the derivative with respect to B (treating A as a constant).
