Let
l e n g h t × w i d t h × h e i g h t = x × y × z lenght\times width\times height=x\times y\times z l e n g h t × w i d t h × h e i g h t = x × y × z Then x y z = 500. xyz=500. x yz = 500.
Let c = c= c = the cost of the square centimeter of the material for the bottom and
the sides. The total cost will be
C = c x y + c ( 2 x + 2 y ) z + 4 c x y C=cxy+c(2x+2y)z+4cxy C = c x y + c ( 2 x + 2 y ) z + 4 c x y
z = 500 x y z=\dfrac{500}{xy} z = x y 500
C = 5 c x y + c ( x + y ) 1000 x y , x > 0 , y > 0 C=5cxy+c(x+y)\dfrac{1000}{xy}, x>0, y>0 C = 5 c x y + c ( x + y ) x y 1000 , x > 0 , y > 0
= 5 c ( x y + 200 x + 200 y ) =5c(xy+\dfrac{200}{x}+\dfrac{200}{y}) = 5 c ( x y + x 200 + y 200 )
∂ C ∂ x = 5 c ( y − 200 x 2 ) \dfrac{\partial C}{\partial x}=5c(y-\dfrac{200}{x^2}) ∂ x ∂ C = 5 c ( y − x 2 200 )
∂ C ∂ y = 5 c ( x − 200 y 2 ) \dfrac{\partial C}{\partial y}=5c(x-\dfrac{200}{y^2}) ∂ y ∂ C = 5 c ( x − y 2 200 ) Find the critical number(s)
∂ C ∂ x = 0 \dfrac{\partial C}{\partial x}=0 ∂ x ∂ C = 0
\dfrac{\partial C}{\partial y}=0\ \
5 c ( y − 200 x 2 ) = 0 5c(y-\dfrac{200}{x^2})=0 5 c ( y − x 2 200 ) = 0
5 c ( x − 200 y 2 ) = 0 5c(x-\dfrac{200}{y^2})=0 5 c ( x − y 2 200 ) = 0 Then
x 2 y = 200 = x y 2 = > x = y , x > 0 , y > 0 x^2y=200=xy^2=>x=y, x>0, y>0 x 2 y = 200 = x y 2 => x = y , x > 0 , y > 0
x − 200 x 2 = 0 x-\dfrac{200}{x^2}=0 x − x 2 200 = 0
x = 200 3 = y x=\sqrt[3]{200}=y x = 3 200 = y
∂ 2 C ∂ x 2 = 5 c ( 400 x 3 ) \dfrac{\partial ^2C}{\partial x^2}=5c(\dfrac{400}{x^3}) ∂ x 2 ∂ 2 C = 5 c ( x 3 400 )
∂ 2 C ∂ y 2 = 5 c ( 400 y 3 ) \dfrac{\partial ^2C}{\partial y^2}=5c(\dfrac{400}{y^3}) ∂ y 2 ∂ 2 C = 5 c ( y 3 400 )
∂ 2 C ∂ x ∂ y = 5 c \dfrac{\partial ^2C}{\partial x\partial y}=5c ∂ x ∂ y ∂ 2 C = 5 c
D = 5 c ( 400 x 3 ) ⋅ 5 c ( 400 y 3 ) − ( 5 c ) 2 D=5c(\dfrac{400}{x^3})\cdot5c(\dfrac{400}{y^3})-(5c)^2 D = 5 c ( x 3 400 ) ⋅ 5 c ( y 3 400 ) − ( 5 c ) 2 If x = y = 200 3 x=y=\sqrt[3]{200} x = y = 3 200
∂ 2 C ∂ x 2 = 5 c ( 400 200 ) = 10 c > 0 \dfrac{\partial ^2C}{\partial x^2}=5c(\dfrac{400}{200})=10c>0 ∂ x 2 ∂ 2 C = 5 c ( 200 400 ) = 10 c > 0
D = 25 c 2 ( 4 − 1 ) > 0 D=25c^2(4-1)>0 D = 25 c 2 ( 4 − 1 ) > 0 The function C ( x , y ) C(x, y) C ( x , y ) has a local minimum at ( 200 3 , 200 3 ) . (\sqrt[3]{200}, \sqrt[3]{200}). ( 3 200 , 3 200 ) .
Since the function C ( x , y ) C(x, y) C ( x , y ) has the only extremum for x > 0 , y > 0 , x>0, y>0, x > 0 , y > 0 , The function C ( x , y ) C(x, y) C ( x , y ) has the absolute minimum at ( 200 3 , 200 3 ) . (\sqrt[3]{200}, \sqrt[3]{200}). ( 3 200 , 3 200 ) .
z = 500 200 3 200 3 = 5 25 3 z=\dfrac{500}{\sqrt[3]{200}\sqrt[3]{200}}=5\sqrt[3]{25} z = 3 200 3 200 500 = 5 3 25 The dimensions of the can must be
l e n g h t × w i d t h × h e i g h t lenght\times width\times height l e n g h t × w i d t h × h e i g h t
= 200 3 c m × 200 3 c m × 5 25 3 c m =\sqrt[3]{200}\ cm\times\sqrt[3]{200}\ cm\times 5\sqrt[3]{25}\ cm = 3 200 c m × 3 200 c m × 5 3 25 c m
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