Suppose you own a tour bus and you book groups of 20 to 70 peo...

Question

Suppose you own a tour bus and you book groups of 20 to 70 people for a day tour. The cost per person is $30 minus $0.25 for every ticket sold. If gas and other miscellaneous costs are $200, how many tickets should you sell to maximize your profit? Treat the number of tickets as a nonnegative real number.

Accepted Answer

Welcome back, everyone. In this problem, a coffee shop sells a special blend of coffee at $15 per pound and offers a discount of 10 cents for each pound purchased. If the fixed costs for producing the coffee are $100 how many pounds should the shop sell to maximize the profit? A says 68 pounds, B 75 pounds, C 78 pounds, and the D says 80 pounds. Now, if we're going to solve this problem, first we need to set up an equation for the profit and then figure out how we can maximize that equation for the profit. And we're gonna use a little bit of, a little bit of differentiation, maximum minimum to do that. But first, let's come up with that equation, OK? Now, let's let X. Let's let X be equal to the number of pounds. Of Cafe Soul OK. The number of pounds of coffee sold. Now if we think about it here, the profit is the revenue from selling the coffee minus the costs. Now what's the revenue? Well, if we think about it here. As our problem says, uh, we're selling the coffee at $15 per pound with a discount of 10 cents for each pound purchased, OK? That means the price per pound is 15 minus 0.10X because of that 10 cents discount, which then means that the revenue is going to be equal to the price per pound times the number of pounds. In other words, X multiplied by 15 minus 0.1 X. And when we expand that, then we get our revenue to be 15X minus 0.1 X squared, OK? No, the fixed cost. I 100, OK. So, our cost function C of X equals 100, which then means that the profit is going to be the revenue minus the cost, OK. So our profit equals R X minus CX, which is gonna be 15 X minus 0.1 X squared minus 100. So, now, we have an equation for our profit. Now that we have the equation for a profit, let's try to maximize the profit by first finding the critical points, OK. Now to find the critical points, we need to take the derivative of the profit function and set it equal to zero. So differentiating the Xs. Eating PRX. Then we should find that its derivative is going to be equal to 15 minus 0.2 X. OK? We, because we can differentiate 15 X and 0, 0.1 X squared with a poor rule, and the derivative of -100 is 0 because it's a constant. No, Thus, if we set that to 0 at the critical point, OK, at the critical points, then 15 minus 0.2 X will be equal to 0, OK, which means then. That here. 0.2 X. Or implies that 0.2 X will be equal to 15, and thus X is going to be equal to 15 divided by 0.2 or 75. So at the critical point, X is equal to 75. Now, we need to make sure that this is the maximum profit and not the minimum. So to test if it's a maximum, we can check the second derivative or analyze the behavior of the first derivative around this point. Let's do it using the second derivative test, OK? For the second derivative. OK. So are given our second derivative of PF X, which means that we're gonna differentiate P X and the derivative of 15, that's gonna be the derivative of 15 minus 0.2 X which equals negative 0.2, OK. Then, since the second derivative is negative, the function is concave down at X equals 75 indicating a maximum, OK. So let's make a note here. The function is concave down. Indicating A maximum, OK. So that means our point where X equals 75 is a maximum point. Thus we know it's the X value for the maximum profit. So what does this mean? Well, therefore, that tells us then that a coffee shop, a coffee shop. Should sell. 75 pounds. Of coffee to maximize their profit. If we look back at our answer traces, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Profit Function

Revenue Maximization

Optimization Techniques

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