# Thread: [SOLVED] i need urgent help with application (word) problems!!!

1. ## [SOLVED] i need urgent help with application (word) problems!!!

hi i'm in grade 11 math and im exceed in failing at application questions. I can do all the regular stuff really easy but twhen it comes to coming up with an equation by myself its really hard for me. So here is the question. Basically what we are learningis taking an quadratic equation and finidng the mins and max value by using the "completing the square" method. heres the actual question

An orchard owner has maintained records that show that, if 25 apple trees are planted in one acre, then each tree yeilds an average 500 apples. The yield decreases by 10 apples per tree for each additional tree that is planted. How many trees should be planted for a maximum yeild.

So here was my reasoning in word form then in an equation

The max yield = the number of trees planted * the amount of apples per tree

I let x represent the additional amount of trees planted because thats whatyou need to know:

y = (25 + x)(500-10x) does this equation not seem proper to you? The only problem with it is that you get some insane number when you multiply it out and complete the square to put it in vertex form. For people who would like to work bacwards here is the answer: 37 or 38 and yes there are to answers which does make sense. Also i could figure this out just by trial and error but id rather not because i need to be able to do this properly. Please to whoever responds, EXPLAIN our answer!!! thanksin adavance

2. well nevermind folks im really sorry but i taslked to my friend and i forgot one vital step that toke me doiwn...i forgot to sub the x back in for the amount of trees i got but whatever the point is that i got it. I giot a new one that is harder and i need help more with. Its somewhat of the same idea as the other one just a little different.

vitaly and jen have 24 m of fencing to enclose a vegetable garden at the back of their house. what are they dimensions of the largest rectagular garden they could encolse with this length of fencing. Now remeber their are only 3 sides of fence because its built against the house so heres 2 equations that i came up with:

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x

P(perimiter) = x + 2y
area = (x)(y)

now where do i go from there lol

it erased all the spaces from my diagram up there so nvm that

3. Hello, Matt!

There are 24 m of fencing to enclose a vegetable garden against a house.
What are the dimensions of the rectagular garden with the largest area?
Code:
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y |                 | y
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x
Here are the two equations:

. . Perimeter: . $x + 2y \:=\:24\;\;[1]$
. . . . . .Area: . . . $A \:=\:xy\;\;\,[2]$

Solve [1] for $x\!:\;\;x \:=\:24- 2y$

Substitute into [2]: . $A \:=\:(24-2y)y$

We have a down-opening parabola: . $A \:=\:24y - 2y^2$

It achieves its maximum at its vertex: . $y \:=\:\frac{-b}{2a}$

We have: . $a \,=\,-2,\;b \,=\,24,\;c\,=\,0$

Hence: . $y \:=\:\frac{-24}{2(-2)} \:=\:6$

Substitute into [1]: . $x + 2(6) \:=\:24\quad\Rightarrow\quad x \,=\,12$

Therefore, the garden should be 12 m long and 6 m high.

4. thanks alot man! I finally get it now, again thanks!