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Bezzdna [24]
1 year ago
12

The annual inventory cost C for a manufacturer is given below, where Q is the order size when the inventory is replenished. Find

the change in annual cost when Q is increased from 340 to 341, and compare this with the instantaneous rate of change when Q = 340. (Round your answer to two decimal places.
Engineering
1 answer:
Nataly_w [17]1 year ago
8 0

The change in annual cost when Q is increased from 340 to 341 is -1.23 and the instantaneous rate of change when Q = 340 is -1.25

<h3>How to find the Instantaneous rate of change?</h3>

The annual inventory cost C for a manufacturer is given as;

C = (1012000/Q) + 7.5Q

where Q is the order size when the inventory is replenished.

Now, the change in C can be calculated by evaluating the cost function at Q = 340 and Q = 341

Change in C = [1,012,000/341 + 7.5*341] - [1,012,000/340 + 7.5*340] ≈ -1.23

Instantaneous rate of change in C is first order derivative C':

C'(Q) = -1,012,000/(Q²) + 7.5

C'(340) = -1,012,000/(340²) + 7.5 ≈ -1.25

Read more about Instantaneous rate of change at; brainly.com/question/14666106

#SPJ1

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6 0
3 years ago
The exhaust steam from a power station turbine is condensed in a condenser operating at 0.0738 bar(abs). The surface of the heat
lozanna [386]

Answer:

Percentage change 5.75 %.

Explanation:Given ;

Given

 Pressure of condenser =0.0738 bar

Surface temperature=20°C

Now from steam table

Properties of steam at 0.0738 bar  

Saturation temperature corresponding to saturation pressure =40°C      

 h_f= 167.5\frac{KJ}{Kg},h_g= 2573.5\frac{KJ}{Kg}

So Δh=2573.5-167.5=2406 KJ/kg

Enthalpy of condensation=2406 KJ/kg

So total heat=Sensible heat of liquid+Enthalpy of condensation

Total\ heat\ =C_p\Delta T+\Delta h

Total heat =4.2(40-20)+2406

Total heat=2,544 KJ/kg

Now film coefficient before inclusion of sensible heat

  h_1=\dfrac{\Delta h}{\Delta T}

  h_1=\dfrac{2406}{20}

h_1=120.3\frac{KJ}{kg-m^2K}

Now film coefficient after inclusion of sensible heat

 h_2=\dfrac{total\ heat}{\Delta T}

 h_2=\dfrac{2,544}{20}

h_2=127.2\frac{KJ}{kg-m^2K}

So\ Percentage\ change=\dfrac{h_2-h_1}{h_1}\times 100

             =\dfrac{127.2-120.3}{120.3}\times 100

                   =5.75 %

So Percentage change 5.75 %.

3 0
2 years ago
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