Solved on Feb 02, 2024

Find the rate of change of sales with respect to advertising spending for the function S(x)=0.002x3+0.7x2+7x+500S(x) = -0.002x^3 + 0.7x^2 + 7x + 500 on the interval 0x2000 \leq x \leq 200. Determine if sales are increasing faster at $110,000\$110,000 or $160,000\$160,000 in advertising spend.

STEP 1

Assumptions
1. The function S(x)S(x) represents the sales in terms of the amount of money xx spent on advertising.
2. The domain of xx is from 00 to 200200, which likely represents thousands of dollars, so the actual money spent on advertising ranges from 00 to 200,000200,000.
3. We are asked to find the second derivative of the sales function, S(x)S^{\prime \prime}(x), which represents the rate of change of the rate of change of sales with respect to the amount of money spent on advertising.
4. We need to compare the second derivative at x=110x = 110 and x=160x = 160 to determine when the total sales are increasing at a faster rate.

STEP 2

Find the first derivative of the sales function S(x)S(x), which will give us the rate of change of sales with respect to the amount of money spent on advertising.
S(x)=ddx(0.002x3+0.7x2+7x+500)S^{\prime}(x) = \frac{d}{dx}(-0.002x^3 + 0.7x^2 + 7x + 500)

STEP 3

Differentiate each term of S(x)S(x) with respect to xx to find S(x)S^{\prime}(x).
S(x)=0.0023x2+0.72x+7S^{\prime}(x) = -0.002 \cdot 3x^2 + 0.7 \cdot 2x + 7

STEP 4

Simplify the expression for S(x)S^{\prime}(x).
S(x)=0.006x2+1.4x+7S^{\prime}(x) = -0.006x^2 + 1.4x + 7

STEP 5

Now find the second derivative of the sales function S(x)S(x), which is the derivative of S(x)S^{\prime}(x), to get S(x)S^{\prime \prime}(x).
S(x)=ddx(0.006x2+1.4x+7)S^{\prime \prime}(x) = \frac{d}{dx}(-0.006x^2 + 1.4x + 7)

STEP 6

Differentiate each term of S(x)S^{\prime}(x) with respect to xx to find S(x)S^{\prime \prime}(x).
S(x)=0.0062x+1.4S^{\prime \prime}(x) = -0.006 \cdot 2x + 1.4

STEP 7

Simplify the expression for S(x)S^{\prime \prime}(x).
S(x)=0.012x+1.4S^{\prime \prime}(x) = -0.012x + 1.4

STEP 8

Evaluate S(x)S^{\prime \prime}(x) at x=110x = 110 to find the rate of change of the rate of change of sales when $110,000 is spent on advertising.
S(110)=0.012110+1.4S^{\prime \prime}(110) = -0.012 \cdot 110 + 1.4

STEP 9

Calculate S(110)S^{\prime \prime}(110).
S(110)=1.32+1.4S^{\prime \prime}(110) = -1.32 + 1.4

STEP 10

Simplify the expression for S(110)S^{\prime \prime}(110).
S(110)=0.08S^{\prime \prime}(110) = 0.08

STEP 11

Evaluate S(x)S^{\prime \prime}(x) at x=160x = 160 to find the rate of change of the rate of change of sales when $160,000 is spent on advertising.
S(160)=0.012160+1.4S^{\prime \prime}(160) = -0.012 \cdot 160 + 1.4

STEP 12

Calculate S(160)S^{\prime \prime}(160).
S(160)=1.92+1.4S^{\prime \prime}(160) = -1.92 + 1.4

STEP 13

Simplify the expression for S(160)S^{\prime \prime}(160).
S(160)=0.52S^{\prime \prime}(160) = -0.52

STEP 14

Compare S(110)S^{\prime \prime}(110) and S(160)S^{\prime \prime}(160) to determine when the total sales are increasing at a faster rate.
Since S(110)=0.08S^{\prime \prime}(110) = 0.08 and S(160)=0.52S^{\prime \prime}(160) = -0.52, and a positive second derivative indicates that the rate of change of sales is increasing, while a negative second derivative indicates that the rate of change of sales is decreasing, we can conclude that Cannon's total sales are increasing at a faster rate when the amount of money spent on advertising is $110,000.
The solution to the problem is that Cannon's total sales are increasing at a faster rate when the amount of money spent on advertising is $110,000.

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