Answer:
<em>perimeter</em><em> </em><em>of </em><em>figure</em><em> </em><em>A </em><em>is </em><em>4</em><em>0</em><em> </em><em>cm</em>
<em>.</em><em>.</em>
<em>breadth </em><em>of </em><em>figure</em><em> </em><em>B </em><em>is </em><em>3</em><em>.</em><em>3</em><em>3</em><em>. </em><em> </em><em>{</em><em>ie,</em><em>3</em><em> </em><em>whole </em><em>4</em><em> </em><em>by </em><em>1</em><em>2</em><em>}</em>
Answer: Weird
Step-by-step explanation: Maybe calculate all the numbers I guess.
Let
x------> the length side of the square base of the box
y-------> the height of the box
we know that
volume of the box=b²*h
b=x
h=y
volume=256 cm³
so
256=x²*y------>y=256/x²--------> equation 1
<span>The amount of material used is directly proportional to the surface area, so we will minimize the amount of material by minimizing the surface area.</span>
surface area of the cardboard=area of the base+perimeter of base*height
area of the base=x²
perimeter of the base=4*x
height=y
surface area=x²+4x*y-----> equation 2
substitute equation 1 in equation 2
SA=x²+4x*[256/x²]-----> SA=x²+1024/x
step 1
find the first derivative of SA and equate to zero
2x+1024*(-1)/x²=0------> 2x=1024/x²----> x³=512--------> x=8 cm
y=256/x²------> y=256/8²-----> y=4 cm
the answer is
the length side of the square base of the box is 8 cm
the height of the box is 4 cm
For the first one it’s D.3
