Question

Hunter and Brian are creating a poster for a project and they need to combine two different size boards to create their project. If the total area cannot exceed 7x to the second power -6x+2 square inches and one board Is 2x to the second power -9x+8 square inches, then what must be the maximum area of the second board?

Answer 1

Answer:

The maximum area of the second board must be $(5x^2+3x-6)$ square inches.

Step-by-step explanation:

The total area of two different size boards cannot exceed $(7x^2-6x+2)$ square inches.

The area of one board is  $(2x^2-9x+8)$ square inches.

Suppose, the area of the second board is  $y$ square inches.

That means......

$y+(2x^2-9x+8)\leq 7x^2-6x+2\\ \\ y\leq (7x^2-6x+2)-(2x^2-9x+8)\\ \\y \leq 7x^2-6x+2-2x^2+9x-8\\ \\ y\leq 5x^2+3x-6$

So, the maximum area of the second board must be $(5x^2+3x-6)$ square inches.

Answer 2

Answer:

5x² + 3x - 6

Step-by-step explanation:

Data:

• Total area: 7x² - 6x + 2

• Area of one board: 2x² - 9x + 8

• Area of the other board: f(x)

The addition of the area of the two boards cannot exceed the total area. Then:

f(x) + 2x² - 9x + 8 ≤ 7x² - 6x + 2

f(x) ≤ 7x² - 6x + 2 - 2x² + 9x - 8

f(x) ≤ (7x² - 2x²) + (-6x + 9x) + (2 - 8)

f(x) ≤ 5x² + 3x - 6 in square inches