Let
x be length of base
y be width of base
z be height of container
then
(1) x=2y
(2) Volume = xyz = 2y^2 * z = 4
(3) Surface area = 2xy + 2yz + 2xz = 4y^2 + 2yz + 4yz [by (1)] = 4y^2 + 6yz = min
Can you finish?
Heres a question from my homework i am struggling with. Hoping someone can explain this to me. Thanks. A rectangular storage container with a lid is said to have a volume of 4m3. the length of its base is twice the width. material for the base costs $2 per m2. material for the sides and lid cost $4 per m2. find the dimensions of the container which will minimize cost and the minimum cost.
Let
x be length of base
y be width of base
z be height of container
then
(1) x=2y
(2) Volume = xyz = 2y^2 * z = 4
(3) Surface area = 2xy + 2yz + 2xz = 4y^2 + 2yz + 4yz [by (1)] = 4y^2 + 6yz = min
Can you finish?
Ok. We can re-write (2) so that z = 2/(y^2) and then substitute this for z into our surface area equation (3). Hence, (3) becomes 4y^2 + 6y(2/(y^2)) = 4y^2 + 12/y = f(y) = min. (3)*
We want to minimize f(y) In order to find the minimum of a one dimensional function, we need to find the critical points through differentiation and then evaluate them, in the interval (0,infinity) as you can't have negative or zero width for the base.
f'(y)= 8y -12/(y^2). Setting this to zero gives us our critical points. 8y - 12/(y^2) = 0 iff 8y^3 - 12 = 0 [we can multiply by y^2 without worrying about multiplying by 0 since y not equal to zero implies y^2 not equal to zero)]
y^3 = 3/2 iff y = (3/2)^(1/3) is our only critical point. plug this value back into (3)* to get the total minimum surface area and then calculate the cost normally. if you need to prove that the only critical point yields the minimum you can test when y < (3/2)^(1/3) and y > (3/2)^(1/3)