Question

select all the correct answers. a farmer wants to build two fenced - off sections within his field, one in the shape of a rectangle and the other in the shape of a square. the side of the square must be equal to the width of the rectangle, x feet. the length of the rectangle must be 50 feet longer than its width. the field the farmer wants to build the two fenced sections in has an area of y square feet. the difference of the area of this field and the area of the fenced, square section needs to be at least 1,000 square feet. in addition, the sum of the fenced areas must be less than the area of the field. this is the system of inequalities that represents this situation. y≥x² + 1,000 y>2x² + 50x which points represent viable solutions? (5,3,000) (30,2,000) (20,2,200) (10,1,100) (25,2,500)

Explanation:

Step1: Substitute point (5, 3000)

For $y\geq x^{2}+1000$, when $x = 5$, $x^{2}+1000=5^{2}+1000=25 + 1000=1025$, and $3000\geq1025$. For $y>2x^{2}+50x$, when $x = 5$, $2x^{2}+50x=2\times5^{2}+50\times5=2\times25 + 250=50+250 = 300$, and $3000>300$.

Step2: Substitute point (30, 2000)

For $y\geq x^{2}+1000$, when $x = 30$, $x^{2}+1000=30^{2}+1000=900 + 1000=1900$, and $2000\geq1900$. For $y>2x^{2}+50x$, when $x = 30$, $2x^{2}+50x=2\times30^{2}+50\times30=2\times900+1500 = 1800+1500=3300$, and $2000<3300$ (not a solution).

Step3: Substitute point (20, 2200)

For $y\geq x^{2}+1000$, when $x = 20$, $x^{2}+1000=20^{2}+1000=400 + 1000=1400$, and $2200\geq1400$. For $y>2x^{2}+50x$, when $x = 20$, $2x^{2}+50x=2\times20^{2}+50\times20=2\times400 + 1000=800+1000 = 1800$, and $2200>1800$.

Step4: Substitute point (10, 1100)

For $y\geq x^{2}+1000$, when $x = 10$, $x^{2}+1000=10^{2}+1000=100 + 1000=1100$, and $1100\geq1100$. For $y>2x^{2}+50x$, when $x = 10$, $2x^{2}+50x=2\times10^{2}+50\times10=2\times100+500 = 200 + 500=700$, and $1100>700$.

Step5: Substitute point (25, 2500)

For $y\geq x^{2}+1000$, when $x = 25$, $x^{2}+1000=25^{2}+1000=625+1000 = 1625$, and $2500\geq1625$. For $y>2x^{2}+50x$, when $x = 25$, $2x^{2}+50x=2\times25^{2}+50\times25=2\times625+1250 = 1250+1250=2500$, and $2500 = 2500$ (not a solution as $y>2x^{2}+50x$ requires strict - inequality).

Answer:

(5, 3000), (20, 2200), (10, 1100)