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# FFPC-INV1 QUIZ

### The Quadratic Relationships (length vs area of rectangles with fixed perimeter)

Question | Answer |
---|---|

A rectangle has a perimeter of 14. Determine the dimensions of ALL of the possible rectangles. | 1x6 4x3 2x5 5x2 3x4 6x1 |

If the perimeter of a rectangle is equal to 50, then the L plus the width is equal to . . . ? | L + W + L + W = 50 so, L + W = 25 (half of the perimeter) |

L + W = 45. Solve the equation for L, in terms of W. | L = 45 - W |

L + W = 250. Solve the equation for W, in terms of L. | W = 250 - L |

If the area of rectangles with a fixed perimeter is A = L(26 - L), determine the fixed perimeter. | 52, the fixed perimeter is double the constant in the formula. |

If the area of rectangles with a fixed perimeter is A = L(78 - L) and the length is 40, determine the width. | In the equation, A = L(78 - L), the quantity (78-L) represents the width so plug in 40 to get 78 - 40 = 38 as the width. |

If the area of rectangles with a fixed perimeter is A = L(100 - L) and the length is 60, determine the area. | A = 60(100 - 60) A = 60(40) A = 2400 |

A rectangle has a fixed perimeter of 8, determine the rectangle with the maximum area. | 1x3 = 3 2x2 = 4 3x1 = 3 , so 2x2 has maximum area |

If a rectangle has a fixed perimeter, the rectangle with the maximum area is always a _____________________ . | Square, L = W |

A rectangle has a perimeter of 32. Determine the dimensions of ALL of the possible rectangles. | 1x15, 2x14, 3x13, 4x12, 5x11, 6x10, 7x9, 8x8 plus all of the rectangles in reverse. |

The following is a list of dimensions of rectangles with a fixed area. Which one represents the rectangle with the maximum area? Explain. 1x15, 2x14, 3x13, 4x12, 5x11, 6x10, 7x9, 8x8 | 8x8, because it is a square plus it's area is 64 which is the largest number. |

Given a graph comparing length and area of rectangles with a fixed perimeter, how can you determine the maximum area? | By looking at the highest point on the graph. |

L + W = 68. Solve the equation for L, in terms of W. | L = 68 - W |

L + W = 190. Solve the equation for W, in terms of L. | W = 190 - L |

Rectangles with a fixed perimeter of 80 have a length of L and a width of 40 - L. Write an equation for Area using only A and L as the variables. | A = L(W) A = L(40 - L), must have parenthesis for full credit |

The table displays rectangles with a fixed perimeter. Length 0 1 2 3 4 5 6 7 8 Area 0 7 12 15 16 15 12 7 0 Determine the fixed perimeter. | 16, Double the length that corresponds to an area of zero. Or determine the width and find the perimeter, ex: L=1 A=7 so W=7, so perimeter is 1 + 7 + 1 + 7. |

The table displays rectangles with a fixed perimeter. Length 0 1 2 3 4 5 6 7 8 Area 0 7 12 15 16 15 12 7 0 Determine the maximum area and the dimensions of that rectangle with the maximum area. | 16 and 4x4, the rectangle with the maximum area is always a square. |

Created by:
gunningta