Break-Even Point Nicolas Inc. sells a product for 63 per...

Break-Even Point Nicolas Inc. sells a product for $63 per unit. The variable cost is $32 per unit, while fixed costs are $284,456. Determine (a) the break-even point in sales units and (b) the break-even point if the selling price were increased to $69 per unit. a. Break-even point in sales units b. Break-even point if the selling price were increased to $69 per unit 9,176 units units Feedback Check My Work a. Unit sales price minus unit variable costs equals unit contribution margin. b. Fixed costs divided by unit contribution margin = break-even point in units.

Transcript

00:01 Hi there, so for this problem, we are told to suppose that a company has a fits cost.

00:07 So the fits cost for this is equal to $300 and a variable cost, and that variable cost is equal to 0 .8 times x plus 1 ,440.

00:22 Now we're told that there is a price, the price for this is 1 ,600 minus 0 .3 times x.

00:29 So then in each order, assume that the number of units must be a whole number.

00:34 Now for part a of this problem, the question is to find the break -even points.

00:40 Okay, so to find the break -even points, first we need to write a cost function and the revenue function.

00:46 Now the cost function is the fits cost, that is $300, and plus the variable cost times x, so we need to multiply all of this by x, so that will give us 0 .8 times x squared plus 1 ,440 times x.

01:01 Now the revenue is just the price times x, so that will be 1 ,600 times x minus 0 .3 times x squared.

01:10 Once we have this, the break -even points is when we set these two expressions equal to the order.

01:15 So we will have 300 plus 0 .8 times x squared plus 1 ,440 times x, and then this is equal to 1 ,600 times x minus 0 .3 times x squared.

01:31 Now we just need to start solving here.

01:34 We can move this to this side, so that will be 0 .38 plus 0 .3, so that will give us 1 .1 times x squared.

01:43 Then we can move this to the other side as well, so that will be 1 ,440 minus 1 ,600, so that will give us minus 160 times x, then this plus 300.

01:56 Now we set this equal to 0, so as you can see in here, this is a quadratic equation, so we can use a quadratic formula to obtain the possible two solutions for this.

02:07 When we do that, we obtain the following two solutions.

02:11 When x, remember that we need to round that to the nearest value, so that will be x equals to 2 and x equals to 144.

02:19 So with this, you can answer.

02:21 Now we can already obtain the revenue for each case.

02:25 For the case of 2, first we are asked about the largest number of units.

02:31 Now for the largest number of units, of course, we know that is 144.

02:38 What you need to do is to evaluate the revenue of this function 144, so that will be the revenue from before, that is 1 ,600 times this value, and that minus 0 .3 times 144 to the square.

02:59 So let's obtain this by using our calculator.

03:08 The value that we obtain is 224 ,179 .2, so that for this value, this is the largest value.

03:18 Now the smallest number is just 2, so we just evaluate that at 2.

03:22 So when x is equal to 2, we obtain 1 ,600 times 2 minus 0 .3 times 2 to the square.

03:31 So let's use our calculator for this, and the value that we obtain is 3 ,198 .8.

03:42 So that's the solution for part a.

03:45 Now for part b, the question is to find the maximum revenue.

03:47 So to maximize the revenue, we derive the revenue with respect to x first.

03:52 So we derive this expression, this with respect to x, that will give us 1 ,600, and then this minus 0 .6 times x.

04:02 We set this equal to 0.

04:04 Now we solve for x in here.

04:06 Then this is the value of x that maximizes the revenue, so that will be 1 ,600 divided by 0 .6.

04:12 So let's use our calculator for this...

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