This would be no solution , there is none
Answer:
Let's call the length of the field "l", and the width of the field "w".
If the area of the field is 72 square meters, then we have:
l x w = 72
And if the length is 6 meters longer than the width, we have:
l = w+6
So looking at the first equation (l x w = 72), we can substitute the l for a w+6.
And we obtain:
(w+6) x (w) = 72
Which simplifies to w^2 + 6w = 72.
This quadratic equation is pretty easy to solve, you just need to factor it.
w^2 + 6w - 72 = 0
(w-6)(w+12)
This leaves the roots of the quadratic equation to be 6 and -12, but in this case, a width of -12 wouldn't make sense.
So, the width of the rectangular field is 6, and the length of the field is 12.
Let me know if this helps!
180-107=73
73 degrees is the measure of the suppplement
Answer:
A = (100 - y)y
Maximum area = 2500 sq. feet.
Step-by-step explanation:
Let the length of the combined rectangular area is y feet and the shared width is x feet.
So, perimeter of the two rectangle together is (3x + 2y) = 200 {Given}
⇒ 2y = 200 - 3x
⇒ 
...... (1).
a) Now, area of the total plot in sq. feet is 
........ (2)
So, this is the expression for area A in terms of length of shared side x.
b) For area to be maximum the condition is 
Now, differentiating equation (2) on both sides with respect to x we get
⇒ x = 33.33 feet.
So, from equation (1) we get
⇒ x = 50 feet.
So, the value of maximum area = 50 × 33.33 = 1666.5 sq. feet. (Answer)
