# Optimization Problem

## Homework Statement

I have been stuck working on this problem for the past little while and can't seem to figure it out.

A construction company needs to create a trucking route for five years to transport ore from a mine site to a smelter. The smelter is located on a major highway 10km from where the mine is 3km off the road.

Construction Costs:
• Upgrade highway costs $200 000/km • Build new gravel road from mine to highway is$500 000/km

Operating conditions:
• There will be 100 return trips each day for 300 days in a year for 5 years
• Operating costs on the gravel road will be $65/h and average speed of 40km/h • Operating costs on the highway will be$50/h and a average speed of 70km/h

Edit: Forgot to mention the book says not to consider the time value of money in your calculations.

## Homework Equations

Pythagorean Theorem: c^2=a^2+b^2

## The Attempt at a Solution

I figured first I need to figure how to construct the route from the mine on the gravel road and highway. Using a Pythagorean I came up with this equation for the cost to build the road: C(x)=500 000$$\sqrt{x^2+3^2}$$+200 000(10-x). Now I'm not sure if I should include operation into the above equation or just make a separate equation all together. I tried to make one big equation into the formula, but I found it confusing and wasn't sure what equation represented. Any help is appreciated.

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What book did this problem come from?

I will try and work it soon, I'm just so curious about where it came from :p

It's from my Calculus textbook. It's from Harcourt Canada and I can't seem to find much more information about it. Seems like the textbook's website has been taken down too.

Edit: The minimum for C(x) I wrote at the beginning I got was about 1.3.

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I tried from the first equation of C(x)=500 000$$\sqrt{x^2+3^2}$$+200 000(10-x) and added the operation costs to that equation. So a I got a new equation:
C(x)=500 000$$\sqrt{x^2+3^2}$$ + 200 000(10-x) + 5(300(200(65($$\frac{\sqrt{x^2+3^2}}{40}$$) + 50($$\frac{(10-x)}{70}$$))))

Which I simplified from just multiplying out and combining like terms to:

C(x)=987 500$$\sqrt{x^2+3^2}$$ + $$\frac{1 500 000}{7}$$(10-x)

The minimum of the new function I calculated to be 0.66, but I'm not sure if this makes senses. The new function adds the total of the operating costs of the 5 years on the road and with the initial construction costs.

D H
Staff Emeritus
I tried from the first equation of C(x)=500 000$$\sqrt{x^2+3^2}$$+200 000(10-x) and added the operation costs to that equation. So a I got a new equation:
C(x)=500 000$$\sqrt{x^2+3^2}$$ + 200 000(10-x) + 5(300(200(65($$\frac{\sqrt{x^2+3^2}}{40}$$) + 50($$\frac{(10-x)}{70}$$))))
Here you are assuming the cost of the trip an empty truck makes in going from the smelter to the mine is equal to the cost of the return trip made a full truck returning from the mine.

Which I simplified from just multiplying out and combining like terms to:

C(x)=987 500$$\sqrt{x^2+3^2}$$ + $$\frac{1 500 000}{7}$$(10-x)
You left out the cost of upgrading the highway here.

The minimum of the new function I calculated to be 0.66, but I'm not sure if this makes senses.

I just assumed the cost for the trips both ways was the same because of the wording of the question. It stated the average speed and I used that to calculate the cost of the trips.

Thanks for the pointing my error in the equation. I redid it so the correct function is:

C(x)=987 500$$\sqrt{x^2+3^2}$$ + $$\frac{2 900 000}{7}$$(10-x)

This is my work calculating the minimum value:

$$C^{'}(x)=\frac{987 500x}{\sqrt{x^2+9}}-\frac{2900000}{7}$$

When $$C^{'}=0$$ is the minimum of the function.

$$0=\frac{987 500x}{\sqrt{x^2+9}}-\frac{2900000}{7}$$

$$0=\frac{6912500x-2900000\sqrt{x^2+9}}{7\sqrt{x^2+9}}$$

$$0=6912500x-2900000\sqrt{x^2+9}$$

$$\frac{6912500x}{2900000}=\sqrt{x^2+9}$$

$$\frac{553x}{232}=\sqrt{x^2+9}$$

$$\frac{305809x^2}{53824}=x^2+9$$

$$305809x^2=53824x^2+484416$$

$$251985x^2=484416$$

$$x^2=\frac{484416}{251985}$$

$$x=1.386$$

So my new answer right now is the minimum is at x=1.386