Respuesta :
Answer:
New dimensions of the floor is approximately 301.25 ft by 181.25 ft
Step-by-step explanation:
The question is incomplete. The complete question should be:
Manufacture wants to enlarger it's floor area 1.5 times that of the current facility. The current facility is 260 ft by 140 ft. The manufacture wants to increase each dimension the same amount. Write the dimensions of the new floor.
Given:
Length of floor = 260 ft
Width of floor = 140 ft
The floor area is increased 1.5 times.
To find the new dimensions of the floor.
Solution:
Original area of the floor = [tex]length\times width= 260\times 140=36400\ ft^2[/tex]
New area = [tex]1.5\times Original\ Area = 1.5\times 36,400=54,600\ ft^2[/tex]
Let the length and width be increased by [tex]x[/tex] ft.
Thus, new length = [tex](260+x)\ ft[/tex]
New width = [tex](140+x)\ ft[/tex]
Area of the new floor can be given as:
⇒ [tex]new\ length\times new\ width[/tex]
⇒ [tex](260+x)(140+x)[/tex]
Multiplying using distribution.
⇒ [tex]x^2+260x+140x+36400[/tex]
⇒ [tex]x^2+400x+36400[/tex]
Thus we can equate this with new area to get the equation to find [tex]x[/tex]
[tex]x^2+400x+36400=54600[/tex]
subtracting both sides by 54600.
[tex]x^2+400x+36400-54600=54600-54600[/tex]
[tex]x^2+400x+18200=0[/tex]
Using quadratic formula:
For a quadratic equation [tex]ax^2+bx+c=0[/tex]
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
For the equation [tex]x^2+400x-18200=0[/tex]
[tex]x=\frac{-400\pm\sqrt{(400)^2-4(1)(-18200)}}{2(1)}[/tex]
[tex]x=\frac{-400\pm\sqrt{232800)}}{2}[/tex]
[tex]x=\frac{-400\pm482.49}{2}[/tex]
[tex]x=\frac{-400+482.49}{2} \ and\ x= \frac{-400-482.49}{2}[/tex]
∴ [tex]x\approx 41.25 \ and\ x\approx-441.25[/tex]
Since length is being increased, so we take [tex]x\approx41.25[/tex]
New dimensions are:
New length [tex]\approx 260\ ft + 41.25\ ft =301.25\ ft[/tex]
New width [tex]\approx 140\ ft + 41.25\ ft =181.25\ ft[/tex]
