Answer:
Length = 2.862 ft
Width = 1.431 ft
Height = 3.05 ft
Explanation:
Let L be lenght, W be width, and H be height of the storage container. Since length is to be twice the width, it means L = 2W. The volumn of the container would also be:
[tex]V = LWH = L(2L)H = 2L^2H[/tex]
SInce the volumn constraint is 50, that means:
[tex]2L^2H = 50[/tex]
[tex]H = \frac{25}{L^2}[/tex]
The cost for the base would be its area times unit price
[tex]C_b = A_bU_b = LWU_b = 2L^2U_b = 2*8L^2 = 16L^2[/tex]
Likewise, the cost for the sides would be
[tex]C_s = A_sU_s = 2(L+W)HU_s = 6LH5 = 30LH[/tex]
We can substitute H from the equation above:
[tex]C_s = 30L\frac{25}{L^2} = \frac{750}{L}[/tex]
Therefore the total material cost
[tex]C = C_b + C_s = 16L^2 + \frac{750}{L} [/tex]
To find the minimum of this function, we can take first derivative then set to 0:
[tex]C^{'} = (16L^2)^{'} + (\frac{750}{L})^{'} = 0[/tex]
[tex] 32L - \frac{750}{L^2} = 0[/tex]
[tex]32L = \frac{750}{L^2}[/tex]
[tex]L^3 = 750/32 = 23.4375[/tex]
[tex]L = 23.4375^{1/3} \approx 2.862ft[/tex]
So W = L/2 = 1.43ft and
[tex] H = \frac{25}{L^2} = 3.05 ft[/tex]
If we take the 2nd derivative and substitute L = 2.862 we would have
[tex]C{''} = 32 + 2\frac{750}{L^3} = 32 + \frac{1500}{23.43} = 96 > 0[/tex]
Hence L = 2.862 would yield the minimum material cost
