Question

The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?

          The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?
        
The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit?
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The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit? A) $1,520,000 B) $1,620,000 C) $1,720,000 D) $1,820,000

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The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit? A. $1,820,000 B. $1,520,000 C. $1,720,000 D. $1,620,000

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The annual revenue and cost functions for a manufacturer of zip drives are approximately R(x) = 520x - 0.02x^2 and C(x) = 160x + 100,000, where x denotes the number of drives made. What is the maximum annual profit? A) $1,720,000 B) $1,820,000 C) $1,620,000 D) $1,520,000

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Transcript

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00:01 Okay, in this problem we have the annual revenue and cost functions for a manufacturer of zip drives are approximately r of x, this is our revenue, is equal to 520x minus .02x squared.
00:18 And then we have our cost function c of x and that's equal to 160x plus 100 ,000.
00:26 Okay, so we're saying what is the maximum annual profit? well, our profit is going to be our revenue minus our cost, right? profit is equal to revenue minus cost.
00:35 So that's going to be equal to 520x minus .02x squared.
00:44 And then we're going to take that and subtract, so minus 160x, i'm going to put that in parentheses, and then plus 100 ,000.
00:54 All right, so what i'm going to do is i'm going to combine these two, i'll have first of all negative .02x, i'm going to put it in ascending order as well, .02x squared, and then 520 minus 160, minus 160, i'm putting that into my calculator, is going to be plus 360x, 360x, and then minus 100 ,000.
01:18 So this is my profit.
01:20 Now, the maximum profit is going to be where it is hitting its peak.
01:27 Okay, so i'm going to do profit prime of x, p prime of x, which means i'm going to take the derivative, and then i'm going to set that equal to zero.
01:36 So that's going to be negative .04x plus 360.
01:42 All right, so i'm going to set that equal to zero, and i'm going to solve for x...
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