# Math Help - Quadratic Word Problems

I don't understand these problems at all..

2b. Melissa plans to put a fence around her garden while leaving a 10 ft. opening to get in and out. If she has 120 ft. of fencing, what is the maximum area this fencing can enclose?

3b. An airline transports an average of 800 passengers a week between Chicago and Springfield; however, the route has become unprofitable. A round-trip fare now costs $300, but a company executive estimates that for each$10 increase in the ticket price, ten passengers will be lost to the competition. If they need revenue of $280,000 per week to break even, can they make it profitable? What is the maximum profit they can make per week? How many customers will they lose? 5b. Tom and Bob are given Batman walkie-talkies for Christmas that claim a range of 5 miles. To test them they go to the corner of Elm and Spruce. Tom begins to walk north on Elm St. at a 3mph pace while Bob walks east on Spruce St. at a 2mph pace. How far did each walk before they faded out? Those 5 separate problems need the following steps: 1. Identify the variable, either through drawing or English statement. 2. Create a quadratic equation that will represent the problem. 3. Solve the quadratic (either roots, vertex, or both) 4. Check using calculator 5. Write a sentence state what you have discovered Help would be appreciated! Thanks 2. Originally Posted by jakej78 I don't understand these problems at all.. 2b. Melissa plans to put a fence around her garden while leaving a 10 ft. opening to get in and out. If she has 120 ft. of fencing, what is the maximum area this fencing can enclose? 3b. An airline transports an average of 800 passengers a week between Chicago and Springfield; however, the route has become unprofitable. A round-trip fare now costs$300, but a company executive estimates that for each $10 increase in the ticket price, ten passengers will be lost to the competition. If they need revenue of$280,000 per week to break even, can they make it profitable? What is the maximum profit they can make per week? How many customers will they lose?

5b. Tom and Bob are given Batman walkie-talkies for Christmas that claim a range of 5 miles. To test them they go to the corner of Elm and Spruce. Tom begins to walk north on Elm St. at a 3mph pace while Bob walks east on Spruce St. at a 2mph pace. How far did each walk before they faded out?

Those 5 separate problems need the following steps:
1. Identify the variable, either through drawing or English statement.
2. Create a quadratic equation that will represent the problem.
3. Solve the quadratic (either roots, vertex, or both)
4. Check using calculator
5. Write a sentence state what you have discovered

Help would be appreciated! Thanks
2b. You can count the 10ft opening as if it were material. So basically, you want to find the maximum area you can enclose given 130 feet of fencing. (The answer should be a square with each side 32.5 ft long.)

3b. Every dollar they increase the price, they lose one customer, so $\text{Profit}=(300+x)(800-x)$. Take the derivative, set it equal to zero, and solve for $x$ to get your maximum value.

5b. Distance between them: $x=\sqrt{(3t)^2+(2t)^2}$. Set $x=5$, solve for $t$, and use that to calculate how far they walked.

3. For 3b, so far I've got
(300+x)(800-x)
x^2+500x+240000=profit
What to do next?

For 5b, I've went as far as
5=13t^2
Next step?

Also, -1.5t² + 18t + 4 = 0 to quadratic formula would be what? If you could show me please..

4. Originally Posted by jakej78
For 3b, so far I've got
(300+x)(800-x)
x^2+500x+240000=profit
What to do next?

For 5b, I've went as far as
5=13t^2
Next step?

Also, -1.5t² + 18t + 4 = 0 to quadratic formula would be what? If you could show me please..
$(300+x)(800-x)=240000+500x-x^2$. Maximum value occurs at $x=\frac{-b}{2a}$. $b=500$; $a=-1$.

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$5=\sqrt{13t^2}$. Just solve for $t$. $t=\frac{5}{\sqrt{13}}$

Now if $\text{distance} = \text{speed}\times\text{time}$, how far did they each walk?

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$a=-1.5$, $b=18$, $c=4$. Just plug into the quadratic formula: $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$