# Calculus optimization problem, what speed will minimize the cost of a certain trip?

Hello I need help from the geniuses out there :). Anyway here is the question.

Brenda drives an 18 wheeler and plans to buy her own truck. The cost function per 100 km is given as C(v)= 0.9 + 0.0016v^2, where v is the speed in kilometers per hour. She plans to travel 1500km and is willing to pay $30/h. Determine the speed that will minimize Brenda's cost for the trip.

I know that I have to take the derivative of the cost function but i get lost after :/ Optimization problems are my weakness unfortunately

### 2 Answers

- 9 years agoFavorite Answer
Keep at it; optimization problems are super important!

Ok so the general rule is that you need to find where your function has a minimum or maximum. Think about the picture of your function. When the graphs changes from increasing to decreasing (or vv), there is always a high point or a low point. The derivative reflects this, because the derivative of a high point or a low point must be 0. Just look at the line tangent to the curve when it hits a max or min. So IN GENERAL, all you have to do is take the derivative and set it to zero, then solve it for x or v or whatever your variable is.

For your problem, it is obvious that you can't just take the derivative of this and then solve for which v gives you zero. That would give you f'(v)=2*.0016*v, which is 0 when v=0. But that's stupid because then she would never go anywehre, so we must need something else. If she never gets anywhere, and she is paying $30 per hour, she will owe someone an infinite amount of money, so she needs to move faster.

The problem also states that she needs to go 1500km, and she'll pay 30/h. So how long will the driver need to drive? Well, that depends on the speed, right? So we need to express that information as a function of the speed. The cost of driving will be C=30*t, where t is the number of hours. But t=1500/v (the time to drive will be the distance divided by the speed). So we can substitute that information in for the cost of driving C=30*t=30*1500/v. Now, the cost function given describes it in terms of cost per 100 km so we'll adjust C to reflect this: C=30*15/v.

You can now combine this information with what was given, and say that the total cost is a function of the velocity given as (T for total): T(v)= 0.9+.0016v^2 + 30*15/v Notice that v is in the denominator of one of the added terms, which tells us that the faster they drive, the less it costs branda, but v^2 is multiplying the other term, so the slower the driver goes, the less it costs That makes sense in real life. So if you find T'(v)=.0016*2-450/(v^2), and solve for when the left hand side, T'(v)=0, you will find the best speed.

- mcpetersLv 44 years ago
Mathematically, if x is the gap between entering into and leaving the river then, Distance below river = sqrt(x^2 + 100 and twenty^2) so entire value = 15 X sqrt(x^2+ 100 and twenty^2) Distance on land = three hundred - x so entire value is 10 X (three hundred-x) it really is close yet not precisely your hypotehsis