Area optimization

Question

A metal wire, 40 cm long, is cut in two and each piece bent to form a square. If the sum of their areas is 58 cm^2, who long is each piece of wire?

Answer 1

can't it be x and 40-x?

 

yes..you can

 

 

 

cut into 2 pieces

length of 1 piece = x

length of second piece = 40-x

 

so when you bent first wire into square  length of every side of square =x/4

so area of square = (x/4)²

 

so when you bent second wire into square length of every side of square =(40-x)/4
so area of square = ((40 -x)/4)²

 

(x/4)² + ((40-x)/4)² = 58
x² /16+ (1600 - 80x + x²)/16 = 58 

x² + 1600 - 80x + x² = 928 __________ multiply by 16
2x² - 80x + 672 = 0
x² - 40x + 336 = 0________ divide by 2
(x - 28)(x - 12) = 0_______ by factoring

 

x= 28  or x=12

 

length of 1 piece = x =12
length of second piece = 40-x =40-12=28

Answer 2

Let x = length of one piece of the wire

Translate:
“metal wire, 40 cm long”
so, 40-x is the length of the other piece of wire

“square. If the sum of their areas is 58 cm^2” means
(x/4)² + ((40-x)/4)² = 58                 [side length is x/4]

The math:
(x/4)² + ((40-x)/4)² = 58
x² + `1600 - 80x + x² = 928              [expand; multiply by 16]
2x² - 80x + 672 = 0                          [get 0 on right side]
x² - 40x + 336 = 0                            [this is easier when divide by 2]
(x - 28)(x - 12) = 0                            [either factor or use quadratic formula]
  
Based on the Multiplicative Property of Zero, x must be either (28) or (12).

So, one length is 12 cm and other length is 28 cm (or one length is 28 cm and the other 12 cm).

Answer 3

Let 4x and 40 - 4x be the length of the wires, then

x² + (10-x)² = 58                          {The sum of the areas is 58 cm²)

Solve for x, we get

x = 3 or x = 7

 

So the lengths of the cut wires are 4(3) = 12 cm and 4 (7) = 28 cm