A rectangular box open at the top is to have volume 32 cubic feet.Find its dimensions if the total surface area is minimum
Answers
The dimensions are 4ft , 4ft, 2ft
Given:
the volume 32 cubic feet
To find :
the dimensions of the box
Solve:
surface area (SA)=2lw+2lh+2hw.
xyz = 32
y = 32 / xz
if the box is square with side length x, the surface area is
a = xy+2yz+2xz
( open at top )
= x^2 + 2z(2x)
= x^2 + 4x(32/(x^2))
= x^2 + 128/x
This will be minimized when the derivative with respect to x is zero.
2x - 128/x^2 = 0
2x^3 - 128 = 0
x^3 = 64
x = 4
A square box that is 4 ft by 4 ft and 2 ft deep will have minimum surface area.
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Answer:
the volume is 32 cubic feet .
Step-by-step explanation:
surface area (SA)=2lw+2Lh+2Hw
xyz=32
a=xy+2yz+2xz
X2+2z(2x)
X2+4x(32/X2)
=X2+128/x
2x-128/X2=0
2x3 -128=0
x3=64
x=4 .......
hence proved the solution is completed mark me as brainliest plz....................