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aksik [14]
3 years ago
6

A rectangular parking area measuring 6000 ft squared is to be enclosed on three sides using​ chain-link fencing that costs ​$5.5

0 per foot. The fourth side will be a wooden fence that costs ​$6 per foot. What dimensions will minimize the total cost to enclose this​ area, and what is the minimum​ cost?
Mathematics
1 answer:
kolezko [41]3 years ago
6 0

Answer:

length x =79.2 ft

width y = 75.75 ft

minimum​ cost P = Rs 1742.41212

Step-by-step explanation:

Let the length of the parking area be 'x', and the width  be 'y'.

Then, we can write the following equations:

Then,  Area of the park: A =x×y = 6000

Now,  Price of the fences P = 2×5.5x + 5.5y + 6y

P = 11x + 11.5y

From the first equation, we have that y = 6000/x

Using this value in the equation for P, we have:

P = 11x + 11.5×6000/x = 11x + 69000/x

To find the minimum of this function, we need to take its derivative and then make it equal to zero:

⇒ dP/dx = 11 - 69000/x^2 = 0

⇒x^2 = 69000/11

x  = 79.20 ft

This is the x value that gives the minimum cost.

Now, finding y and P, we have:

x×y = 6000

y = 6000/79.2 = 75.75 ft

P = 11x + 69000/x = Rs 1742.41212

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Suppose a manufacturer finds that 95% of their production is normal but the final 5% has one or more flaws. Each flawed good has
RUDIKE [14]

Answer:

1)    

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW        0.01             0.95

2) 0.04 and $0.04

3) 0.025 and $0.025

4) 0.015 and $0.015

5) 0.95 and $0.95

Step-by-step explanation:

Given that;

financial cost = $1

p(flaw) = 0.05  

p(type 1 flaw / flaw) = 80% = 0.8

p(type 2 flaw / flaw) = 50% = 0.5

p( type 1 and 2 flaw/flaw) = 30% = 0.30

1) Bivariate Table

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

p( only 1 flow) = 0.04 - 0.015 = 0.025

p( only 2 flow) =  0.025 - 0.015 = 0.01

THEREFORE  the Bivariate Table;

FLAW                         TYPE2         NO TYPE2 FLAW

TYPE1                         0.015           0.025

NO TYPE1 FLAW       0.01              0.95

2) probability and expectations of type 1 flaw?

p( type 1 flaw) = p(flaw) × p(type 1 flaw/flaw) = 0.05 × 0.8 = 0.04

Expected financial cost to the firm per good = $1 × 0.04 = $0.04

3)  probability and expectation of Type 2 flaw

p( type 2 flaw) = p(flaw) × p(type 2 flaw/flaw)  = 0.05 × 0.5 = 0.025

Expected financial cost to the firm per good = $1 × 0.025 = $0.025

4) probability and expectations of Type 1 and 2 flaws

p( type 1 and 2 flaw) =  p(flow) × p( type 1 & 2 flaw/flaw) = 0.05 × 0.3 = 0.015

Expected financial cost to the firm per good = $1 * 0.015 = $0.015

5) probability and expectations of no flaws?

Probability of no flaw = P(No flaw) =95% =  0.95

Expected financial cost saved the firm per good due to no flaw

= $1 × 0.95 = $0.95

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