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3 years ago

6

A manufacturer has fixed costs of $16,000 per month and can produce w widgets at a cost of 0.1w^(2) + 20w. How many widgets shou

ld be produced monthly to minimize the cost per widget?

1 answer:

musickatia [10]3 years ago

5 0

Consider c as the cost of the widget so that our given equation is
c = 0.1w^2 + 20w
Take the derivate of the equation.
d/dt (c = 0.1w^2 + 20w)
dc/dt = 0.2w + 20
Given dc/dt = $16000 per month, the number of widgets would contain:
16000 = 0.2w + 20
-0.2w = 20 - 16000
-0.2w = -15980
w = 79900 widgets

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4 years ago

A point H is 20m away from the foot of a tower on the same horizontal ground. From the point H, the angle of elevation of the po

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Answer:

a. See Attachment 1

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c. $HT = 31.1\ m$

d. $OH = 28.4\ m$

Step-by-step explanation:

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To calculate PT, we need to get distance OT and OP

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We have to consider angle 30, distance OH and distance OP

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--------------------------------------------------------

Calculating the distance between H and the top of the tower

This is represented by HT

HT can be calculated using Pythagoras theorem

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Substitute 20 for OH and $OT = 23.836$

$HT^2 = 20^2 + 23.836^2$

$HT^2 = 400 + 568.154896$

$HT^2 = 968.154896$

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$HT = \sqrt{968.154896}$

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--------------------------------------------------------

Calculating the position of H

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See Attachment 2

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The relationship between these parameters is;

$tan50 = \frac{OH}{OT}$

Multiply both sides by OT

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$23.836 * 1.1918 = OH$

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3 years ago