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Helen [10]
2 years ago
10

Champion Bakers uses specialized ovens to bake its bread. One oven costs $689,000 and lasts about 4 years before it needs to be

replaced. The annual operating cost per oven is $41,000. What is the equivalent annual cost of an oven if the required rate of return is 13 percent?
A. ​-$272,638
B. ​-$248,313
C. ​-$232,407
D. ​-$200,561
E. ​$196,210
Business
1 answer:
Strike441 [17]2 years ago
5 0

Answer:

A. ​-$272,638

Explanation:

Firstly, we need to calculate the total present value of initial investment plus all operating cost of the project. Then we will use that amount to solve for equal-amount annual operating cost.

Total present value of the project is -689,000 - 41,000/(1+13%) - 41,000/(1+13%)^2 + ... + 41,000/(1+13%)^4, or -810,953.32 (negative sign mean cost).

So, the equivalent annual cost of an oven is -272,638.

The equation solving problem can be made easily using Excel or BA II plus calculator.

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Read 2 more answers
An oil refinery is located on the north bank of a straight river that is 3km wide. A pipeline is to be constructed from the refi
Mariulka [41]

Answer:

$6,598,076.21

Explanation:

<h2>THE KEY IS TO FIND OUT THE COST FUNCTION, the calculations are very easy!!!</h2><h2></h2><h3>In order to find the cost function, take a look at the drawing attached. </h3>

We can see the river (sort of) that is 3 km wide and the storage tanks on the other side of the river 8 km apart.

<h3 />

Laying pipes under (across) the river costs 1,000,000 the km & laying pipes over land costs 500,000 per km.

<h3 /><h3>So basically the cost function is 1,000,000 multiplied by something plus 500,000 multiplied by another something.</h3><h3 />

The distance across the river can be found by using Pythagoras Theorem. A side is 3 km the other is unknown, so we call it X. And it is equal to:

\sqrt{3^{2} +x^{2}}=\\\sqrt{9 +x^{2}}

And we multiply it by 1,000,000; the cost of laying pipe under the river, the we get:

1000000\sqrt{9+x^{2}

The distance over the land is (8-x), as we can see in the drawing. So we multiply it by its cost, 500,000. And we get 500,000(8-x).

So the cost function f(x) would be:

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)

<h2>From here, we just have to differentiate and the derivative found must be equal to zero in order to minimize cost. </h2><h3>The value of x when the derivative is zero is plugged in the original function to get the cost.</h3><h3 /><h2>LET'S DO THIS</h2>

f(x)=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(x)=1000000(9+x^{2})^{1/2}+4000000-500000x\\f'(x)=\frac{1}{2} 1000000(9+x^{2})^{-1/2}(2x)-500000\\\\f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000

<h2>f'(x)=0</h2>

f'(x)=\frac{1000000x}{\sqrt{9+x^2}}  - 500000=0\\\frac{1000000x}{\sqrt{9+x^2}}  = 500000\\\frac{2x}{\sqrt{9+x^2}}  = 1\\2x={\sqrt{9+x^2}}\\4x^2=9+x^2\\3x^2=9\\x^2=3\\x=\sqrt{3} \\

And we plug square root of 3 in the original cost function  ad we get

f(\sqrt{3} )=1000000\sqrt{9+x^{2}} + 500000(8-x)\\f(\sqrt{3})=1000000\sqrt{9+(\sqrt{3} )^{2}} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{9+3} + 500000(8-(\sqrt{3}))\\f(\sqrt{3})=1000000\sqrt{12}+500000(6.27)\\f(\sqrt{3})=1000000(3.46)+500000(6.27)\\f(\sqrt{3})=3464101.62+3133974.60\\f(\sqrt{3})=6598076.21\\

<h2>so the minimal cost is $6,598,076.21</h2><h2 /><h3 />

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3 years ago
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