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Ross-Simons Company has a monthly advertising budget of $60,000$. Their marketing department estimates that if they spend $x$ dollars on newspaper advertising and $...

Question

Ross-Simons Company has a monthly advertising budget of $60,000$. Their marketing department estimates that if they spend $x$ dollars on newspaper advertising and $y$ dollars on television advertising, then the monthly sales will be given by $z=f(x, y)=90 x^{1 / 4} y^{3 / 4}$$ dollars. Determine how much money Ross-Simons should spend on newspaper ads and on television ads each month to maximize its monthly sales.

Ross-Simons Company has a monthly advertising budget of $60,000$. Their marketing department estimates that if they spend $x$ dollars on newspaper advertising and $y$ dollars on television advertising, then the monthly sales will be given by $z=f(x, y)=90 x^{1 / 4} y^{3 / 4}$$ dollars. Determine how much money Ross-Simons should spend on newspaper ads and on television ads each month to maximize its monthly sales.



Answers

The total revenue $ R $ (in millions of dollars) for a company is related to its advertising expense by the function

$ R = \frac{1}{100,000}(-x^3 + 600x^2) $, $ 0 \le x \le 400 $

where $ x $ is the amount spent on advertising (in tens of thousands of dollars). Use the graph of this function, shown in the figure on the next page, to estimate the point on the graph at which the function is increasing most rapidly. This point is called the point of diminishing returns because any expense above this amount will yield less return per dollar invested in advertising.

This problem about advertising costs is a great one for the graphing calculator because these numbers air really not realistic to do by hand. So what we can do is go into y equals and weaken type the Prophet Equation X is the advertising expense in tens of thousands of dollars and p or why in this case is the average is the profit in dollars. Okay, so then we're told to let X be from 0 to 60. So for when? For the window dimensions, I let my ex minimum equals zero and my ex maximum equal 60. And then I was looking at the rest of the problem and it was talking about a profit of 2.5 $1,000,000. So I decided to let my wide minimum be negative $1 million my wife, Max Mamby, $3 million. So he received chunk of the graph. And what we want to do is now use this to find the smaller to advertising amounts that will give us a profit of 2,000,002 and 1/2 $1,000,000. So what we can do is go back to buy equals and go toe y two and we want to type 2.5 $1,000,000 in there, and that will give us a horizontal line at that height. And we're looking for the two points where those graphs intersect so we can go into the calculate menu. Second trace choose Number five Intersect. And yes, we're in the first part. Yes, we're on the second curve. And then let's move the cursor closer to that first guess, and that gives us 38.4, and then the other one is going to be a larger number. And the problem asks us for the smaller of the two so we could say approximately 38 times 10,000. So $380,000. Or, if you want to get more specific, you could say 3843 Let's see, what would that be? $384,355 or something like that. But, of course, we're not gonna be able to get it to that level of specificity. So just approximately $380,000

For this problem. It will be great to use a graphing calculator. We have a profit equation when a type that into my y equals menu and then we're going to look at that for X values between zero and 50. So we goto window type in zero and 50 as our X Men and X max, and we're looking for the amount for acts that will yield a profit. Why of $800,000? So I just let my graph go as low as negative $1 million as high as $2 million we can look at that portion of the graph. So we're interested in a profit of 800,000 so we can go back to y equals goto y two and type in 800,000. That will put a horizontal line at that value, and we can see that that line intersects the curve and two points, and we want the smaller of the two X values. So we want the the intersection point further to the left. So we go into the calculate menu, which is second Trace, which was number five Intersect Cursors on the first curve. Yes, it's on the second curve? Yes, and now we give it a guess. And the intersection point is approximately 31.5. Then we multiply that by $10,000 so we get $315,000.

And this problem were given that a study done by an advertising agency reveals that when X thousands of dollars are spent on advertising, results in the sales increase in thousands of dollars given by the function as of X equals negative one third times X minus X squared, plus thirty five for X between zero and six. So here X and SFX are measured in thousands of dollars. In part, they were asked to find increasing sales when no money is spent on advertising. So no money corresponds to X equals zero. So as a zero is equal to negative one third time, zero minus X squared plus thirty five, which is equal to twenty three. And so in thousands of dollars, the increase in sales is twenty three thousand dollars. In Part B, we're asked to find the increase in sales when six thousand dollars are spent on advertising, so six thousand dollars spent corresponds tow X equals six for measuring X and thousands of dollars as of six negative one third time, six minus six squared plus thirty five, which is equal to thirty five. So the increase in sales is thirty five thousand dollars. In part C. We need to sketch the graph of SX without a calculator. So to begin, let's plot the points we know in part A. We found that when X equals zero, SFX is twenty three and in part B, we found when X is equal to six. S of X is equal of thirty five. So it's maybe mark out to ten. Over here, there's five. Maybe let's go up to fifty, ten, twenty, thirty, forty, fifty, thirty, forty, fifty. And then negative ten. Over here. So we have the points zero comma twenty three and six comma thirty five. Now, another thing to note is that this graph is just a translation of the graph negative one third X squared. So I'm going to go ahead and graph that this is the graph of negative one third X squared and we see if we compare with negative one third times X minus six squared plus thirty five, the X minus six means we shift right. Six units and the plus thirty five means we shift up thirty five units. So if we take our graph from before we shift it right six and up thirty five, then we'LL get this graph here, there's the graph of our quadratic equation. Negative one third times X minus six squared, plus thirty five

Section 3.4 problem number 50. So we're given in this one a profit function. So p of X is equal to 500 plus 48 X minus four x squared. So this is the prophet and hundreds of dollars. And what we're asked to do is to find the marginal profit. So the marginal profit, it's just gonna be p prime of X. Okay, soapy prime of X p prime of X is the limit is H approaches zero off. So that's P of X plus h. So that's gonna be 500 plus 48 x plus H minus four X plus h squared minus P of x. So that's gonna be minus 500 minus 48 x plus four x squared. All of that over h. So I could see immediately the five hundreds cancel. I can also see a cancellation negative 48 x and the positive 48 x So this becomes the limit as H approaches zero of 48 h minus four X squared and then minus eight x h and then minus four h squared and then plus four x squared over h. And what do I see in a cancellation. Now I see four x squared and minus for X squared. And I also see that this is there's an h in every term here, So h h h that becomes, um, only one h. So this becomes the limit. As H approaches zero of 48 minus eight x minus four h when h goes to zero, this is 48 minus eight x. So I find that the marginal profit is 48 minus eight x. And now we want to evaluate this marginal profit at $400 at $600 at $800 huh? And at $1000 and see what kind of information this gives us. So we need to figure out what is p of four. Because it was already given in hundreds of dollars. So that's 48 um, minus eight times four. So 48 p f four is 48 miners. Attempts for that's 48 minus 32 so that it's gonna be eight monies to a six performances. 16. Okay, so that value is 16. If I look a p prime of six, that's gonna be 48 minus eight times six. That is zero. Okay. And if you look at 800 p, Prime at eight is going to be 48 minus eight times eight, which is 64. So 48 minus 64 is going to be minus 16. And if you look at 1000 that's p prime 10. That's gonna be 48 minus 80. Okay? And so that's going to be what, minus 32. So what does it tell me? It says if I look at this, decide in each case whether or not I should increase my expenditure. So if you look at it, if if my expenditure is $400. Okay, so my expenditures $400. The marginal rates says I'm increasing profits at that ping. So this represents an increase of profits. That's 600 that represents profits. Staying steady. Okay, So if I could spend 400 have an increase of profit, why would I go to 600? Not necessarily. If you look at 808 100 tells me I would have a decrease of profit 10 I would have an even higher decrease of profit. So in that case, it tells me that well, in the best case 400 would be the best of all of these scenarios. You see here now, another way that you can think about this is the original was what, um, 500 plus 48 x minus four x squared on this four X square. We know that the graph of that thing that is a parabola that opens downward. So that's a problem that opens downwards. It tells me that at some point, you know, if I look at that, I've got positive increases and then a zero, and then it looks like I've got negative increases. So it tells me that when I look at slopes of that and derivatives, I expect it to go up a level off zero and to go back down again. So in this case, we looked at the marginal rates from this original function that we had and we found in the four cases, there was one case where that made sense because I had increasing profits at 400. But once I got to 600 that would not be the case. Okay,


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