Answer to Question #87957 in Calculus for MONAAMBIGHAI

Question

Apol is a company that sells smartphone.it relies on two companies: x and y to help manufacture some components in smartphone. the cost in RM to manufacture some components in the smartphone given by following model,
C (x,y)=e^x (32x-16x^2-7)+2ye^x-e^{2x}-y^2
Based on mathematical model above, find extreme points. Then, analyse the relation between the companies and the cost to produce some components in smartphone by apol

Answer 1

$C(x,y)=e^x (32x-16x^2-7)+2ye^x-e^{2x}-y^2$

i) Let's find extreme points:

$\frac{\partial C(x,y)}{\partial x}=e^x (32x-16x^2-7)+e^x (32-32x)+2ye^x-2e^{2x}=e^x (32x-16x^2-7+32-32x)+2ye^x-2e^{2x}=e^x (-16x^2+25)+2ye^x-2e^{2x}$

$\frac{\partial C(x,y)}{ \partial y}=2e^x-2y$

$\begin{cases} e^x (-16x^2+25)+2ye^x-2e^{2x}=0 \\ 2e^x-2y=0 \end{cases}$

$\begin{cases} e^x (-16x^2+25)+2ye^x-2e^{2x}=0 \\ y=e^x \end{cases}$

$e^x (-16x^2+25)+2e^{2x}-2e^{2x}=0$

$e^x (-16x^2+25)=0$

$-16x^2+25=0$

$x^2=25/16$

$x_1=5/4; \; x_2=-5/4$

$y_1=e^{5/4}; \; y_2=e^{-5/4}$

$C(5/4,e^{5/4})=e^{5/4} (32 \cdot 5/4-16(5/4)^2-7)+2e^{5/4}e^{5/4}-e^{5/2}-e^{5/2}=e^{5/4} (40-25-7)=8e^{5/4}$

Set $x_2​=-5/4; \; y_2=e^{-5/4}$ has no meaning as far as number of components can not be negative.

ii) Let's classify obtained extreme point $x​=5/4; \; y=e^{5/4}$

$A=\frac{\partial ^2 C(x,y)}{\partial x^2}=e^x (-16x^2+25)-32xe^x+2ye^x-4e^{2x}=e^x (-16x^2-32x+25)+2ye^x-4e^{2x}$

$B=\frac{\partial ^2 C(x,y)}{\partial x \partial y}=2e^x$

$E=\frac{\partial ^2 C(x,y)}{\partial y^2}=-2$

$A=e^{5/4} (-16(5/4)^2-32 \cdot 5/4+25)+2e^{5/4}e^{5/4}-4e^{5/2}=e^{5/4} (-25-40+25)-2e^{5/2}=-40e^{5/4}-2e^{5/2}$

$B=2e^{5/4}$

$E=-2$

$D=AE-B^2=80e^{5/4}+4e^{5/2}-4e^{5/2}=80e^{5/4}$

$A<0$ and $D>0$, so function $C(x,y)$ has a minimum at $x=5/4; \; y=e^{5/4}$.

Finally, we can conclude that Apol can minimize its costs to the level of RM $8e^{5/4}$ by the taking of 5/4 components from the x company and $e^{5/4}$ components from the y company.