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A company manufactures and sells x video phones per week. The weekly price-demand and cost equations are p = 550 - 0.5x and C(x) = 415x + 10000

Below are the answers I came up. Please tell me if I am correct.

To maximize their weekly revenue, the company should charge $275 per phone and produce 550 phones. The maximum weekly revenue is $151250.

The maximum weekly profit is $-887.5 and occurs when the company produces 135 phones per week at a price of $482.5 per phone.

Thanks!

2007-02-27 16:52:18 · 3 answers · asked by Marla L. 1 in Science & Mathematics Mathematics

3 answers

I get the same answers as you, including the negative profit. Evidently the company will make a loss no matter how many phones it makes, so it shouldn't even be in business.

To be on the safe side, I recommend that you double check to make sure you didn't copy down the problem wrong. Usually math problems are designed so that you don't get negative profits for your optimal solution. If it turns out that you copied down the problem correctly, then your answer is indeed correct.

2007-02-27 18:28:45 · answer #1 · answered by Bramblyspam 7 · 0 0

It all looks OK to me except the negative sign in the figure for the maximum weekly profit. Why do you describe it as a complex calculus problem?

2007-02-27 17:48:46 · answer #2 · answered by Anonymous · 0 0

a) nicely, acceleration is the spinoff or slope of the speed graph, so v'(t) is comparable to a(t). working backwards, you will discover (utilising the regulations of derivatives) that v(t)= -32t ft/s + ninety six ft/s.... you may desire to upload the +ninety six ft/s by way of inital speed. utilising the comparable logic, speed is the spinoff of the placement graph so x'(t)= v(t), artwork backward from this and additionally you be responsive to that place is x(t)= -16t^2 + 96t. b) remedy x(t)= /16t^2 + 96t = 0 for t, then divide that answer by utilising 2. (The t you stumble on is the completed time that the object is in the air.) you may desire to discover that t= 6 seconds, so the time this is at optimum top is 3 seconds. c) Use the respond from b-- plug it into the unique x(t) equation to discover the optimum top reached. The max top is one hundred forty four ft. d) This answer is the 1st cost for t you chanced on partly b... in simple terms set x(t) equivalent to 0 and remedy for t-- we earlier chanced in this that t=6 seconds. e) to discover this, use the function for speed with the completed time the object is in the air. v(6) = -32 ft/s * 6 s + ninety six ft/s= -ninety two ft/s

2016-10-02 02:36:04 · answer #3 · answered by launer 4 · 0 0

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