Asmall resort is situated on an island that lies exactly 5 miles from p, the nearest point to the island along a perfectly straight shoreline. 10 miles down the shoreline from p is the closest source of fresh water. if it costs 1.4 times as much money to lay pipe in the water as it does on land, how far down the shoreline from p should the pipe from the island reach land in order to minimize the total construction costs?

  5.103 miles

Step-by-step explanation:

Let x represent the distance from P down the shoreline. Then the relative construction cost is ...

  c = 1.4×(distance to shore) + 1.0×(10 -x)

where ...

  distance to shore = √(x² +5²)

The derivative of the cost is zero at the minimum, so we have ...

  c = 1.4√(x²+25) +10 -x

  dc/dx = 0 = 1.4x/√(x²+25) -1

  1.4x = √(x² +25) . . . . add 1, multiply by the denominator

  (1.4x)² = x² +25 . . . . . square

  0.96x² = 25 . . . . . . . .subtract x²

  x = 5/√0.96 ≈ 5.103104

The pipeline from the island should reach 5.103 miles down the shoreline from P.

Comment on this solution

If you consider this solution carefully, you may realize that the general solution to this sort of problem is ...

  distance from P = (island-to-P distance)/√(k² -1)

where k is the cost factor for water relative to shoreline, 1.4 in this problem.

You may also realize that the minimum value of k that makes a difference in where the pipe comes ashore is √(1+((island-to-P)/(P to terminal))²). In this problem, that value is √1.25 ≈ 1.11803. For values of k below that, the cheapest route is direct from the island to the terminal (water source).

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step-by-step explanation:

the x's and the y's both go up the same amount each time, therefore it will be a linear function because it will go in a straight line when drawn on a graph.

hope this !

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204