How Can the Quadratic Formula Help Solve Real-World Problems?

[travishillier]

Quadratic formula help ...

Heres the question ... your help is greatly needed and appreciated ...

Markita wants to fence a rectangular plot of land along side the shore of a lake. Only 3 sides must be fenced, since the lake will form teh lake will form teh fourth side. Markita had 100m of fencing, and she wants the plot of land to have an area of 500m^2 (squared). Find teh dimensions of the plot of land, to the nearest tenth of a metre. Expalin and justify your solution.


Theres thequestion , lmk what u can do to help me out ... Thx

 

[cookiemonster]

Please don't post the same topic twice...

cookiemonster

 

[travishillier]

srry won't happen again .. thx

 

[Inspector Gadget]

Well, here's your formulas.

P: 100 = 2x + y [normally, perimeter would be 2x + 2y, but because one of the sides is the lake, you don't count one of the y's]
A: 500 = x*y

Now, solve the first equation for y...

y = 100 - 2x

And substitute it into the area formula...

500 = x(100 - 2x)
500 = 100x - 2x^2
2x^2 - 100x + 500 = 0

Now, you can take the above and put it into the quadratic formula...

[-b +/- sqrt(b^2-4ac)]/2a
a = 2
b = -100
c = 500

Substituting in, you get 44.3 and 5.6 as the two solutions. Now, that's x...now we need y...go back to the perimeter formula...

y = 100 - 2x

Substituting in the two numbers we found above for x, you get...

Solution 1: x = 44.3, y = 11.4
Solution 2: x = 5.6, y = 88.8

If you multiply them to check the area, you'll get around 497 and 505, which isn't exactly 500, but you were asked to give rounded values to the nearest tenth, so that's okay.