A metal box with a square base and vertical sides isbti contain 1024cm³ of water. The material for the top an bottom cost rs 5 per cm² and material for the sidescost rs2.50cm² find following if l=x b=x and h=y then find
1. equation of volume.
2. find the cost of top and bottom.
3. find the cost of whole box. 4. Find the least cost of box
Answers
Answer:
Answer
Since volume of the box =1024cm
3
let length of the side of square base be x cm and height of the box be y cm
∴ volume of box (V) =x
2
×y=1024
Since x
2
y=1024⇒y=
x
2
1024
let C denotes the cost of the box
∴C=2x
2
×5+4xy×2.50
=10x
2
+10xy=10x(x+y)
=10x(x+
x
2
1024
)
=
x
10
(x
3
+1024)
⇒C=10x
2
+
x
10240
...(i)
On differentiating both sides w.r.t. x we get
dx
dC
=20x+10240(−x)
−2
dx
dC
=20x−
x
2
10240
...(ii)
Now,
dx
dC
=0
⇒20x=
x
2
10240
⇒20x
3
=10240
⇒x
3
=512=8
3
⇒x=8
Again differentiating eq (ii) w.r.t. x we get
dx
2
d
2
C
=20−10240(−2).
x
3
1
dx
2
d
2
C
=20+
x
3
20480
At x=8
∴(
dx
2
d
2
C
)
x=8
=20+
512
2080
=60>0
For x=8 cost is minimum and the corresponding least cost of the box is:
C(8)=10.8
2
+
8
10240
=640+1280=1920
∴ least cost =Rs 1920