The maximum grain for corn is achieved by planting at a density of 40,000 plants per acre. A farmer wants to maximize the yield for the field represented on the coordinate grid. Each unit on the coordinate grid represents one foot. How many corn plants, to the nearest thousand, does the farmer need? ( hint 1 acre =43,560 square) The farmer needs approximately about how many corn plant

514,400 plants.

Step-by-step Explanation:

From the attached diagram given below which shows the coordinate grid, we can say the shape of the farm is that of a trapezium.

Number of plants to be planted by the farmer at 40,000 plants per acre = area of land in acres * 40,000.

Let's find the area of the land(trapezium):

Area of trapezium = ½(a+b)*h

From the coordinate grid, from point E to H on the y-axis, we have 900 units = 900 ft (Given that 1 unit = 1 foot).

Also, from F to G, we have 500 units = 500ft; and the height of the trapezium is represented by the number of units the trapezium covers on the x-axis = 800units = 800ft.

a = 900ft, b = 500ft, c= 800ft

==> Area of Land/trapezium = ½(a+b)*h

= ½(900+500)*800

= ½(1400)*800

= 700*800 = 560,000ft²

Given that 1acre = 43,560ft²

x acre = 560,000ft²

x = 560,000*1 /43,560

x = 12.86 acres

Therefore, number of plants to be planted = 12.86 acres * 40,000 plants = 514,400 plants.

i think the answer is b.

step-by-step explanation:

not 100% sure though.