urgent help! 10 easy points to the first to answer correctly??? and 5 stars!?!?!?!!?

helpppp meee pleasse!?!?!??!

I dontt get this!?!?!?!

thanks in advance!!!!!

5. A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. He will use one side of his house as a side of the play area. What dimensions should he use if he wants to maximize the play area?

a.) 10 ft by 40 ft

b.)15 ft by 30 ft

c.) 18 ft by 24 ft

d.) 30 ft by 30 ft

9 Answers

  • Alright, so. We know that one side of the house is going to be used for one side of the rectangle. What that means is that the other three sides have to equal 60 feet when added together.

    A- 10+10+40=60 Okay, so this one might work. Let's put A aside for now.

    B- 15+15+30 = 60. This one might work too! Let's also put this aside for now.

    C- 18+18+24 = 60 This one might work as well! Let's put this aside.

    D- 30+30+30=90 This one won't work because you need more than 60 feet of fence. D is not your answer.

    (The adding was me trying to get 60 feet out of the remaining 3 sides of the rectangle, in case you didn't follow)

    So, the biggest area will be the area with the largest square footage, which we get by multiplying.

    10 X 40 = 400 square feet.

    15 X 30 = 450 square feet. <----this one is the largest.

    18 X 24 = 432 square feet.

    Your answer is B.

  • B

    15 x 30 = 450sq. ft. thats bigger than 10 by 40 (=400) and uses all the fencing.

    15 + 15 + 30 + 0 (0 because the house is the other side of the enclosure) = 60

    people saying D because its the biggest answer don't realise that 30 by 30 requires 4 x 30 ft of fencing, which this man doesn't have AND 30 by 30 ISN'T a rectangle! it's a square!

  • Can't see how d) could mathematically work as 2 x 30 + 30 = 90!!

    a) Area is 10 x 40 = 400 sq ft

    b) Area is 15 x 30 = 450 sq ft

    c) Area is 18 x 24 = 432 sq ft

    so, c) is the maximum area.

    In general, if length l and width w,

    then P = l + 2w = 60 and A = l x w

    => A = (60 - 2w) x w = 60w - 2w^2

    dA/dW = 60 - 4w = 0

    then, w = 15 => l = 30

    i.e. maximum area when the length is twice the width!!

  • Play Area Fencing

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  • You calculate area by using sides given in the answers and whose area will be highest will be dimensions of the play area




    d)30*30=900 and d is you answer

  • The answer is d.

    It is just saying, multiply each number by the other,

    eg. 10 x 40 = 400

    so you can get the biggest area for his pets to play in.

    30 x 30 = 900

    so therefor is the biggest area 🙂

    hope it helps!!

  • b

  • b. i got it somehow but its ok

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