Question

(a) Differentiate the following function
y = x^2 e^x
(b) A cosmetic company is planning the introduction of a promotion of a new lipstick line. The marketing research department after test marketing the new line in a carefully selected city found that the demand in the city is approximately given by p = 12e^(-x), where x which should be within this range, 0 ≤ x ≤ 2 were thousand lipsticks sold per week at a price of Kenya shillings. At what price will the weekly revenue be at maximum? What is the maximum weekly revenue

Answer 1

Answer on Question #51480 – Math - Differential Calculus | Equations

(a) Differentiate the following function

$$y = x ^ {\wedge} 2 e ^ {\wedge} x$$

(b) A cosmetic company is planning the introduction of a promotion of a new lipstick line. The marketing research department after test marketing the new line in a carefully selected city found that the demand in the city is approximately given by $p = 12e^{\wedge}(-x)$, where $x$ which should be within this range, $0 \leq x \leq 2$ were thousand lipsticks sold per week at a price of Kenya shillings. At what price will the weekly revenue be at maximum? What is the maximum weekly revenue

Solution.

(a) $y = x^{2}e^{x}, y' = 2xe^{x} + x^{2}e^{x} = x(x + 2)e^{x}$.

(b) $R = p * x = 12xe^{-x}$;

$$\frac {d R}{d x} = 1 2 e ^ {- x} - 1 2 x e ^ {- x} = 1 2 (1 - x) e ^ {- x}.$$ $$\frac {d R}{d x} = 0 \rightarrow x = 1.$$ $$R (1) = 1 2 e ^ {- 1} \approx 4. 4 1 6.$$

The weekly revenue will be at maximum at price 1 Kenya shilling.

The maximum weekly revenue 4416 Kenya shillings.

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