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mixas84 [53]
3 years ago
10

A construction firm is evaluating two value-adding projects. The first project deals with building access roads to a new termina

l at the local airport. The second project is to build a parking garage on a piece of land that the firm owns adjacent to the airport. The firm's decision will be to(A) Accept both projects because they are independent projects(B) Accept both projects because they are contingent projects(C) Pick the one that adds the most value because they are mutually exclusive projects(D) Pick neither project
Business
2 answers:
Lelu [443]3 years ago
6 0

Answer:

In this case scenario when a firm has two different projects, the firm will

(A) Accept both projects because they are independent projects

Explanation:

The construction firm has to make some decisions right now, it has two candidate projects to add value. The first one is projecting which means building roads of access to the new airport terminal. While the second one is to build a parking garage in a place close to the airport. The firm should pick both because they are not the same project and there is not going to be any problem with the contest for winning the project. In some cases influence matters are involver, but not in this one.

Phantasy [73]3 years ago
4 0

<em>Answer:</em>

<em>The answer is simple. The correct option is  </em><em>accept both projects because they are independent projects.</em>

<em></em>

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A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by upper c equals 0.
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Answer:

The cost per month is increasing at a rate $365.

Explanation:

Differentiation Formula

  • \frac{d}{dx}(x^n)= nx^{n-1}  
  • \frac{d}{dx}(a)=0             [ where a is a constant]
  • \frac{d}{dx}(ax^n)=a \frac{d}{dx}(x^n)= anx^{n-1}

Given that,

A manufacturer of handcrafted wine racks has determined that the cost to produce x units per month is given by

c=0.2x^2+10,000.

Again given that,

the rate of changing production is 13 unit per month

i.e \frac{dx}{dt}=13

To find the cost per month, we need to find out the value \frac{dc}{dt} when production is changing at the rate 13 units per month and the production is 70 units.

c=0.2x^2+10,000

Differentiating with respect to t

\frac{d}{dt}(c)=\frac{d}{dt}(0.2x^2)+\frac{d}{dx}(10,000)

\Rightarrow \frac{dc}{dt}=0.2\frac{d}{dt}(x^2)+\frac{d}{dx}(10,000)

\Rightarrow \frac{dc}{dt}=0.2\times 2x^{2-1}\frac{dx}{dt}+0

\Rightarrow \frac{dc}{dt}=0.4x\frac{dx}{dt}

Plugging \frac{dx}{dt}=13

\Rightarrow \frac{dc}{dt}=0.4x\times 13

\Rightarrow \frac{dc}{dt}=5.2x

\frac{dc}{dt}|_{x=70}=5.2\times 70 [ plugging x=70]

            =364

[ The unit of c is not given. Assume that the unit of c is dollar.]

The cost per month is increasing at a rate $365.

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