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

Studdy Solution

STEP 1

Assumptions
1. The function $S(x)$ represents the sales in terms of the amount of money $x$ spent on advertising.
2. The domain of $x$ is from $0$ to $200$ , which likely represents thousands of dollars, so the actual money spent on advertising ranges from $0$ to $200,000$ .
3. We are asked to find the second derivative of the sales function, $S^{\prime \prime}(x)$ , 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 $x = 110$ and $x = 160$ to determine when the total sales are increasing at a faster rate.

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

Compare $S^{\prime \prime}(110)$ and $S^{\prime \prime}(160)$ to determine when the total sales are increasing at a faster rate.
Since $S^{\prime \prime}(110) = 0.08$ and $S^{\prime \prime}(160) = -0.52$ , 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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Question Find the rate of change of sales with respect to advertising spending for the function $S(x) = -0.002x^3 + 0.7x^2 + 7x + 500$ on the interval $0 \leq x \leq 200$ . Determine if sales are increasing faster at $\$110,000$ or $\$160,000$ in advertising spend.

Studdy Solution

STEP 1

STEP 2

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

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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Studdy Solution

STEP 1

STEP 2

STEP 3

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

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

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

STEP 10

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

STEP 11

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

STEP 12

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

STEP 13

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

STEP 14

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

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

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

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

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

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

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

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

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

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

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