use differential approximations in the following problem. a company will sell n units of a product after…

use differential approximations in the following problem. a company will sell n units of a product after spending $x thousand in advertising, as given by n = 50x - x², 5≤x≤25. approximately what increase in sales will result by increasing the advertising budget from $15,000 to $16,000? from $20,000 to $21,000? find the differential dn. dn = ( ) dx the increase in sales from increasing the advertising budget from $15,000 to $16,000 is approximately units. (type a whole number.) the increase in sales from increasing the advertising budget from $20,000 to $21,000 is approximately units. (type a whole number.)

use differential approximations in the following problem. a company will sell n units of a product after spending $x thousand in advertising, as given by n = 50x - x², 5≤x≤25. approximately what increase in sales will result by increasing the advertising budget from $15,000 to $16,000? from $20,000 to $21,000? find the differential dn. dn = ( ) dx the increase in sales from increasing the advertising budget from $15,000 to $16,000 is approximately units. (type a whole number.) the increase in sales from increasing the advertising budget from $20,000 to $21,000 is approximately units. (type a whole number.)

Answer

Explanation:

Step1: Differentiate N with respect to x

Given $N = 50x - x^{2}$, using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{dN}{dx}=50 - 2x$. Then $dN=(50 - 2x)dx$.

Step2: Calculate increase for $x = 15$

When $x$ increases from $15$ to $16$, $dx=1$. Substitute $x = 15$ into $dN=(50 - 2x)dx$. So $dN=(50-2\times15)\times1=(50 - 30)\times1 = 20$.

Step3: Calculate increase for $x = 20$

When $x$ increases from $20$ to $21$, $dx = 1$. Substitute $x = 20$ into $dN=(50 - 2x)dx$. So $dN=(50-2\times20)\times1=(50 - 40)\times1=10$.

Answer:

$dN=(50 - 2x)dx$ $20$ $10$