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y 600- 0 Area 2(mr) Area (2wrh 10 Video Example) EXAMPLE 2 A cylindrical can is to be made to hold 600 cm³ of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in the figure, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From the figure, we see that the sides are made from a rectangular sheet with dimensions 2πr and h. So the surface area is A = 2πr²² + 3 To eliminate h we use the fact that the volume is given as 600 cm³. Thus Th=600 which gives h = Α = 2πμ + 2πη( = 2πr²2² + . Substitution of this into the expression for A gives A (r) = Therefore the function we want to minimize is = = 2πr² + A'(r) = 4πr - 4( To find the critical numbers, we differentiate: Then A'(r) = 0 when 1200 T ) r>0 the absolute minimum. = 300, so the only critical number is r = Since the domain of A is (0, ∞), we can't use the endpoint arguments to determine if the critical point is a maximum or a minimum. But we can observe that A'(r) < 0 for r < and A'(r) > 0 for r > , so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus r = must give give

y 600- 0 Area 2(mr) Area (2wrh 10 Video Example) EXAMPLE 2 A cylindrical can is to be made to hold 600 cm³ of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in the figure, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From the figure, we see that the sides are made from a rectangular sheet with dimensions 2πr and h. So the surface area is A = 2πr²² + 3 To eliminate h we use the fact that the volume is given as 600 cm³. Thus Th=600 which gives h = Α = 2πμ + 2πη( = 2πr²2² + . Substitution of this into the expression for A gives A (r) = Therefore the function we want to minimize is = = 2πr² + A'(r) = 4πr - 4( To find the critical numbers, we differentiate: Then A'(r) = 0 when 1200 T ) r>0 the absolute minimum. = 300, so the only critical number is r = Since the domain of A is (0, ∞), we can't use the endpoint arguments to determine if the critical point is a maximum or a minimum. But we can observe that A'(r) < 0 for r < and A'(r) > 0 for r > , so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus r = must give give

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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[Alternatively, we could argue that A(r) → ∞ as r → 0˚ and A(r) → ∞ as r → ∞, so there must be a minimum value of A(r), which must occur at the critical number. See the graph.] The value of h corresponding to r = ³√300/π is h = 600 TR² 600 = 2r Thus, to minimize the cost of the can, the radius should be ³√300/7 cm and the height should be equal to twice the radius, namely, the diameter.
Transcribed Image Text:[Alternatively, we could argue that A(r) → ∞ as r → 0˚ and A(r) → ∞ as r → ∞, so there must be a minimum value of A(r), which must occur at the critical number. See the graph.] The value of h corresponding to r = ³√300/π is h = 600 TR² 600 = 2r Thus, to minimize the cost of the can, the radius should be ³√300/7 cm and the height should be equal to twice the radius, namely, the diameter.
y 600- 0 =1 Area 2(mr) Area (2 wr)h 10 Video Example h EXAMPLE 2 A cylindrical can is to be made to hold 600 cm³ of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in the figure, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From the figure, we see that the sides are made from a rectangular sheet with dimensions 2πr and h. So the surface area is A = To eliminate h we use the fact that the volume is given as 600 cm³. πρη = = 600 which gives h = A = 2πr² + = 2πr² + = 2πι2 + 2πι( Substitution of this into the expression for A gives Therefore the function we want to minimize is 1200 A (r) = 2πr² + r> 0 Then A'(r) To find the critical numbers, we differentiate: A'(r) = 4ær · 4( 2 = 0 when T ) = 300, so the only critical number is r = Since the domain of A is (0, ∞), we can't use the endpoint arguments to determine if the critical point is a maximum or a minimum. But we can observe that A'(r) < 0 for r < and A'(r) > 0 for r > the absolute minimum. Thus so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus r= must give give
Transcribed Image Text:y 600- 0 =1 Area 2(mr) Area (2 wr)h 10 Video Example h EXAMPLE 2 A cylindrical can is to be made to hold 600 cm³ of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can. SOLUTION Draw the diagram as in the figure, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). From the figure, we see that the sides are made from a rectangular sheet with dimensions 2πr and h. So the surface area is A = To eliminate h we use the fact that the volume is given as 600 cm³. πρη = = 600 which gives h = A = 2πr² + = 2πr² + = 2πι2 + 2πι( Substitution of this into the expression for A gives Therefore the function we want to minimize is 1200 A (r) = 2πr² + r> 0 Then A'(r) To find the critical numbers, we differentiate: A'(r) = 4ær · 4( 2 = 0 when T ) = 300, so the only critical number is r = Since the domain of A is (0, ∞), we can't use the endpoint arguments to determine if the critical point is a maximum or a minimum. But we can observe that A'(r) < 0 for r < and A'(r) > 0 for r > the absolute minimum. Thus so A is decreasing for all r to the left of the critical number and increasing for all r to the right. Thus r= must give give
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