Answer to Question #275633 in Calculus for ron

Question

A manufacturer packages his product in a 500-ml cylindrical can. The material for the

top costs four times as much per square centimeter as that to be used for the bottom and

the sides. What must be its dimensions so that the cost of the can is least?

Answer 1

Let

$$lenght\times width\times height=x\times y\times z$$

Then $xyz=500.$

Let $c=$ the cost of the square centimeter of the material for the bottom and

the sides. The total cost will be

$$C=cxy+c(2x+2y)z+4cxy$$

$$z=\dfrac{500}{xy}$$

$$C=5cxy+c(x+y)\dfrac{1000}{xy}, x>0, y>0$$

$$=5c(xy+\dfrac{200}{x}+\dfrac{200}{y})$$

$$\dfrac{\partial C}{\partial x}=5c(y-\dfrac{200}{x^2})$$

$$\dfrac{\partial C}{\partial y}=5c(x-\dfrac{200}{y^2})$$

Find the critical number(s)

$$\dfrac{\partial C}{\partial x}=0$$

$$\dfrac{\partial C}{\partial y}=0\ \$$

$$5c(y-\dfrac{200}{x^2})=0$$

$$5c(x-\dfrac{200}{y^2})=0$$

Then

$$x^2y=200=xy^2=>x=y, x>0, y>0$$

$$x-\dfrac{200}{x^2}=0$$

$$x=\sqrt[3]{200}=y$$

$$\dfrac{\partial ^2C}{\partial x^2}=5c(\dfrac{400}{x^3})$$

$$\dfrac{\partial ^2C}{\partial y^2}=5c(\dfrac{400}{y^3})$$

$$\dfrac{\partial ^2C}{\partial x\partial y}=5c$$

$$D=5c(\dfrac{400}{x^3})\cdot5c(\dfrac{400}{y^3})-(5c)^2$$

If $x=y=\sqrt[3]{200}$

$$\dfrac{\partial ^2C}{\partial x^2}=5c(\dfrac{400}{200})=10c>0$$

$$D=25c^2(4-1)>0$$

The function $C(x, y)$ has a local minimum at $(\sqrt[3]{200}, \sqrt[3]{200}).$

Since the function $C(x, y)$ has the only extremum for $x>0, y>0,$ The function $C(x, y)$ has the absolute minimum at $(\sqrt[3]{200}, \sqrt[3]{200}).$

$$z=\dfrac{500}{\sqrt[3]{200}\sqrt[3]{200}}=5\sqrt[3]{25}$$

The dimensions of the can must be

$$lenght\times width\times height$$

$$=\sqrt[3]{200}\ cm\times\sqrt[3]{200}\ cm\times 5\sqrt[3]{25}\ cm$$