Lindsie is painting on a canvas with dimensions of 3 ft by 4 ft. she wants to make a display model for a gallery that has twice the area of the original. how much should she increase the length and width by, if she wants to increase them by the same amount? round your answer to the nearest tenth.
B. 1.4 feet
Let, the amount of increase be 'x' ft.
Since, the length and width of the canvas are 4 ft and 3 ft respectively.
Thus, area of the canvas, = length × breadth = 4 × 3 = 12 ft²
Since, the area of display model is twice the area of the canvas. We have,
= 2 ×
i.e. = 2 × 12
i.e. = 24 ft².
As, the length and width of the canvas are increased by 'x'.
The, length and width of the display model are (x+4) ft and (x+3) ft.
So, we get,
= length × breadth = (x+4) × (x+3) =
Since, = 24 ft²
i.e. = 24
Solving the quadratic equation, we get,
i.e. x = -8.4 and x= 1.4
Since, the value of x cannot be negative.
Thus, x = 1.4 feet.
B. 1.4 feet
Lindsie is painting on a canvas of 3 ft by 4 ft. She wants to make a display model for a gallery that has twice the area of the original .
The painting on the canvas has the dimension 3 ft by 4 ft . The area can be calculated as follows :
since it is a rectangle,
area = Lb
area = 3 × 4
area = 12 ft²
The new gallery display will have twice the area of the painting on canvas
12 × 2 = 24 ft²
If she wants to increase the length and the width by the same amount the product of the new width and length will be equal or closest to 24 ft²
let x be the same value we are adding
(x + 3)(x + 4) = 24
x² + 4x + 3x + 12 = 24
x² + 7x + 12 = 24
x² + 7x -12 = 0
using Almighty formula
(-b ±√b² - 4ac)/2a
a = 1
b = 7
c = -12
(-7 ± √49 + 48)/2
(-7 ± √97)/2
(-7 ± 9.8488578018 )/2
(-7 + 9.8488578018) /2 0r (-7 - 9.8488578018) /2
2.8488578018 /2 or -16.8488578018 /2
1.4244289009 ft or 8.4244289009 ft
x = 1.4244289009 ft , 8.4244289009 ft
x = 1.4244289009 ft is the best value
x ≈ 1.4 ft
Hope this helps.