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

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.

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

5 0

3 years ago

Find the distance between (-5,6)and (3,2).

skad [1K]

Answer:

$\displaystyle d = 4\sqrt{5}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Coordinates (x, y)

Algebra II

Distance Formula: $\displaystyle d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Step-by-step explanation:

Step 1: Define

Point (-5, 6) → x₁ = -5, y₁ = 6

Point (3, 2) → x₂ = 3, y₂ = 2

Step 2: Find distance d

Simply plug in the 2 coordinates into the distance formula to find distance d

Substitute in points [Distance Formulas]: $\displaystyle d = \sqrt{(3+5)^2+(2-6)^2}$
[Distance] [√Radical] (Parenthesis) Add/Subtract: $\displaystyle d = \sqrt{(8)^2+(-4)^2}$
[Distance] [√Radical] Evaluate exponents: $\displaystyle d = \sqrt{64+16}$
[Distance] [√Radical] Add: $\displaystyle d = \sqrt{80}$
[Distance] [√Radical] Simplify: $\displaystyle d = 4\sqrt{5}$