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Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

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BuyFindarrow_forward

Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

Consider the situation in Exercise 51 if the cost of laying pipe under the river is considerably higher than the cost of laying pipe over land ($400,000/km). You may suspect that in some instances, the minimum distance possible under the river should be used, and P should be located 6 km from the refinery, directly across from the storage tanks. Show that this is never the case, no matter what the “under river” cost is.

To determine

To show: That P cannot be located directly across from the storage tank.

Explanation

Given:

The cost of the laying pipe over land = $400000/km

The cost of the laying pipe under the river = $400000/km

Formula used:

Pythagorean Theorem:

If c denotes the length of the hypotenuse and a and b denote the length of the other two sides of a right angle triangle, then the Pythagorean theorem can be expressed as the Pythagorean equation: a2+b2=c2

Proof:

Let the cost of the laying pipe under the river be $k00000/km>$400000/km.

Let P be (6x) km east to refinery.

In Figure 1,

Consider the triangle ΔPAS,

By Pythagorean Theorem,

PS=x2+22=x2+4

P is x2+4 far from Storage tank.

Hence, the length of the pipeline on the land is (6x).

The cost for pipeline on the land is 400000(6x).

The length of the pipeline under the river is x2+4.

The cost for pipeline on the land is k00000x2+4

The total cost is c(x)=k00000x2+4+400000(6x)

Differentiate c(x) with respect to x,

c(x)=k00000×12×2xx2+4400000=k00000×xx2+4400000

For critical points, c(x)=0,

k00000×xx2+4400000=0kxx2+4=4kx=4x2+4k2x2=16x2+64

Simplify further as follows,

x2(k216)=64x=±8(k216)

Distance cannot be negative, hence x=8(k216)

Differentiate c(x) with respect to x,

c(x)=k00000×x2+4x×12×2xx2+4(x2+4)

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