A certain product has daily fixed costs of 180 and variable...

A certain product has daily fixed costs of $180 and variable costs of $30 per unit. The selling price, $p$, for the product is $p = 58 - x$. (a) The total costs to make 25 units of the product is $ $ No $ sign. No comma. No space. (b) The maximum revenue is $ $ (Hint: $R(x) = px$) No $ sign. No comma. No space. (c) The maximum profit is $ No $ sign. No comma. No space.

Transcript

00:01 All right, we've got a couple of functions here.

00:03 We have a cost and revenue functions.

00:06 Let's see, the cost of revenue and give the value represents the number of units.

00:13 We have a revenue function, r of x, which is equal to negative 0 .25 x squared plus 202 .5x.

00:25 And then we have a cost function.

00:27 And that is equal to 97 .5x plus, what is that? 111 .01 8 .75.

00:40 So if x represents the number of units, our revenue function tells us how much money we're going to make as we sell that number of units.

00:50 But our cost function tells us how much it costs us to make that product.

00:54 So now you're going to profit.

00:56 What is the profit? well, profit in general is going to be able to do how much money you're making minus cost.

01:05 How much do it cost you to make that number? so we just need to take these functions, negative 0 .25x squared plus 2 .02 .5x.

01:16 And we need to subtract 97 .5x plus 11 ,000, 18755.

01:28 All right, so we'll just add these together.

01:33 Negative 0 .25x squared.

01:36 No like terms to add with that.

01:38 Now don't forget here, we have to distribute this negative.

01:44 That's why the depramphease.

01:46 So 202 .5 minus 97 .5.

01:51 That gives us 105.

01:52 So we end up with plus 105x minus minus that 11 ,01875.

02:01 So there's our profit function, this is p of x.

02:04 It's how much profit we're going to make for each unit we sell.

02:08 Okay, the next part of the question, how many items must be sold to maximize the revenue? so we're going to go back to our revenue function up here.

02:17 We've got r of x.

02:18 And what we need, what we're basically looking for, so if we put this on a graph, this would, it's a quadratic, it's going to graph as a parabola.

02:25 So what we're basically looking for is the vertex.

02:29 If we find the vertex, that's going to give us the maximum point.

02:32 You know, this is a negative here, which means it's going to be a parabola that opens down.

02:37 We're going to have a maximum point.

02:39 So to find the x value of the vertex, you can say x is equal to negative b over 2a.

02:47 And so that's going to be equal to negative 202 .5 over 2 times negative negative.

02:55 0 .25.

02:59 And that's equal to 405.

03:00 405.

03:01 So how many items must be sold to maximize the revenue? 405 items...

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