Respuesta :
Answer:
[tex]2m\times 2m\times \frac{5}{4}m[/tex]
Step-by-step explanation:
We are given that a container with square bottom.
Let side of square=x
Height of container=h
Volume of container=[tex]5m^3[/tex]
Cost of 1 square meter of material for the bottom=$10
Cost of 1 square meter of material for side=$8
We have to find the dimension of least expensive container.
Volume of container=[tex]x^2h[/tex]
[tex]x^2h=5[/tex]
[tex]h=\frac{5}{x^2}[/tex]
Surface area of container=[tex]x^2+2(x+x)h=x^2+4xh[/tex]
Cost=[tex]C(x)=10x^2+8(4xh)=10x^2+32x(\frac{5}{x^2})=10x^2+\frac{160}{x}[/tex]
[tex]C(x)=10x^2+\frac{160}{x}[/tex]
Differentiate w.r.t x
[tex]\frac{dC}{dx}=20x-\frac{160}{x^2}[/tex]
[tex]\frac{dC}{dx}=0[/tex]
[tex]20x-\frac{160}{x^2}=0[/tex]
[tex]\frac{160}{x^2}=20x[/tex]
[tex]x^3=\frac{160}{20}=8[/tex]
[tex]x=\sqrt[3]{8}=2[/tex]
Because side of container is always positive.
Again differentiate w.r.t x
[tex]\frac{d^2C}{dx^2}=20+\frac{320}{x^3}[/tex]
Substitute x=2
[tex]\frac{d^2C}{dx^2}=20+\frac{320}{2^3}=60>0[/tex]
Hence, the cost is minimum at x=2
Substitute the value of x
[tex]h=\frac{5}{2^2}=\frac{5}{4}[/tex]m
Hence, the dimensions of the least expensive container are
[tex]2m\times 2m\times \frac{5}{4}m[/tex]
Answer:
1.26 m , 3.15 m
Step-by-step explanation:
Let the side of the square base is y.
Height of the container is h.
Area of base = y x y = y²
Area of side walls = 4 x y x h = 4yh
Volume of the container, V = y²h
According to the question, the volume of the container is 5 m³
So, 5 = y²h
h = 5 / y²  .... (1 )
Cost of bottom, C1 = 10 y²
Cost of side walls, C2 = 8 yh
Total cost, C = 10 y² + 8yh
Substitute the value of h from equation (1)
C = 10 y² + 8 y x 5 / y²
C = 10y² + 40 / y
Differentiate it with respect to y
dC/dy = 20 y - 40/y²
For maxima and minima, dC/dy = 0
20 y = 40/y²
y³ = 2
y = 1.26 m
So, h = 5 / (1.26 x 1.26) = 3.15
Thus, the side of base is 1.26 m and height is 3.15 m to minimize the cost.
