Given information:
A rectangular storage container with an open top is to have volume of 10 m3 with length of its base is twice the width. The base costs of material is $10 per square meter and the sides costs of material is $6 per square meter.
Formula used:
Let l be the length w be the width and h be the height of cuboid.
I. The lateral surface area of cuboid A=2wh+2hl
II. Volume of open box V=lwh
And
Let f be a differentiable function defined on an interval I and let a∈I .
Then
a) x=a is a point of local maximum value of f, if
i) f′(a)=0 and
ii) f′(x) changes sign from positive to negative as x increases through a , i.e. if f′(x)>0 at every point sufficiently close to and to the left of a , and f′(x)<0 at every point sufficiently close to and to the right of a , then a is a point of local maxima
b) x=a is a point of local maximum value of f, if
i) f′(a)=0 and
ii) f′(x) changes sign from negative to positive as x increases through a , i.e. if f′(x)<0 at every point sufficiently close to and to the left of a , and at f′(x)>0 every point sufficiently close to and to the right of a , then a is a point of local minima.
c) f′(a)=0 and If f′(x) does not change sign as x increases through a , then a is neither a point of local maxima nor a point of local minima.
d)
i. If f″(a)>0 then f has a local minimum at x=a
ii. If f″(a)<0 then f has a local maximum at x=a
Calculation:
As per the given problem
The rectangular storage container with an open top is to have volume of 10 m3 with length of its base is twice the width
Let width=w Then, lenght(l)=2w
Recall that, Let l be the length w be the width and h be the height of cuboid.
Volume of open box V=lwh
Substitute the values, and solve for h :
2w×w×h=10 w2×h=5 h=5w2 ......(1)
Let l be the length w be the width and h be the height of cuboid
.
The lateral surface area of cuboid A=2wh+2hl
The base costs of material is $10 per square meter and the sides costs of material is $6 per square meter.
Total cost(C)=10×Area of base of cuboid +6×Area of lateral surface area of cuboid =10×lw+6×(2wh+2hl)
Substitute l=2w and h=5w2 and further simplify,
C′(w)=120.14≥0 at Total cost(C)=10×lw+6×(2wh+2hl) =10×2w×w+6×2×w×5w2+6×2×5w2×2w =20w2+60w+120wTotal cost(C)=10×lw+6×(2wh+2hl) =10×2w×w+6×2×w×5w2+6×2×5w2×2w =20w2+60w+120w C(w)=20w2+180w
Recall that,
Let f be a differentiable function defined on an interval I and let a∈I .
Then
a) x=a is a point of local maximum value of f, if
i. f′(a)=0 and
ii. f′(x) changes sign from positive to negative as x increases through a , i.e. if f′(x)>0 at every point sufficiently close to and to the left of a , and f′(x)<0 at every point sufficiently close to and to the right of a , then a is a point of local maxima
b) x=a is a point of local maximum value of f, if
i. f′(a)=0 and
ii. f′(x) changes sign from negative to positive as x increases through a , i.e. if f′(x)<0 at every point sufficiently close to and to the left of a , and at every point sufficiently close to and to the right of a , then a is a point of local minima.
c) f′(a)=0 and If f′(x) does not change sign as x increases through a , then a is neither a point of local maxima nor a point of local minima.
d)
If f″(a)>0 then f has a local minimum at x=a
If f″(a)<0 then f has a local maximum at x=a
Differentiate on both sides,
C′(w)=20×2w−180w2C′(w)=40w−180w2 ......(2)
Solve for C′(w)=0 ,
40w−180w2=040w=180w2w3=18040w=923w=1.65(approx)
Differentiate equation (2) with respect to w
C′(w)=40+2×180w3C′(w)=40+360w3
Substitute w=1.65(approx) and simplified
C′(1.65)=40+2×180(1.65)3C′(w)=40+3601.65×1.65×1.65C′(w)=40+80.14C′(w)=120.14(appox)
Therefore,
C′(w)=120.14≥0 at w=1.65
For w=1.65 volume of container is minimize
Substitute w=1.65 in equation (1)
h=5(1.65)2h=51.65×1.65h=1.84(approx)
And
l=10whl=101.65×1.84l=3.29(approx)
The total cost(C)=10×lw+6×(2wh+2hl) =10×3.29×1.65+6×(2×1.65×1.84+2×1.84×3.29) =163.54(approx)
The cost of material for the cheapest such container is $163.54(approx)