Answer:
![Area = 1690ft^2\\x = 130ft, y = 26ft](https://img.qammunity.org/2020/formulas/mathematics/college/p7k1fpvkmt6ledb3e33o2dfeo1s4la8tg6.png)
Explanation:
Given the costs we can form an equation:
![10y + 2x =520 - Eq(A)](https://img.qammunity.org/2020/formulas/mathematics/college/fhq3spuviaacs933gswv84f2fsuluuc1z0.png)
and fencing is triangular such that the area enclosed can be written as:
![A = (xy)/(2) -Eq(B)](https://img.qammunity.org/2020/formulas/mathematics/college/rqf3jj1nsd9i8kf4908y8kfivxopg95aw0.png)
- First need to convert the above equation so that it is only in terms of one variable. [either x or y]
To make the equation only in terms of
we can substitute
from Eq(A) i.e,
, to Eq(B)
![A = (x)/(2) (520-2x)/(10)](https://img.qammunity.org/2020/formulas/mathematics/college/vnjqz0lnnbjeqjqzb94zmkfml6sbwomdup.png)
simplify
![A = (520x - 2x^2)/(20)](https://img.qammunity.org/2020/formulas/mathematics/college/xu3p6d6h3gfk92ohvnmykh1sv60go3ilbf.png)
![A = -(1)/(10)x^2 + 26x](https://img.qammunity.org/2020/formulas/mathematics/college/n2acvh12htwfi8kwn25cah157ee65edw3v.png)
- Now, in order to find the maximum area enclosed we can find
and equate it zero.
![(dA)/(dx) = -(1)/(5)x + 26](https://img.qammunity.org/2020/formulas/mathematics/college/5t5kuc62gnlim0xcow65evmr8f0smksp6x.png)
![0 = -(1)/(5)x + 26](https://img.qammunity.org/2020/formulas/mathematics/college/wiw53qnbgytg8g9g749qmf86gr0vk16gfz.png)
![-26 = -(1)/(5)x](https://img.qammunity.org/2020/formulas/mathematics/college/t7uocd57khzehtc5oh9tau3qter8709m71.png)
![x = 130](https://img.qammunity.org/2020/formulas/mathematics/college/o35hclhwowgtgt53uhpyqufzk3qqsiplkd.png)
we have the length of one dimension: specifically, the lower fence will be
![x =130ft](https://img.qammunity.org/2020/formulas/mathematics/college/jd9lt6qfmx38bvf0u1i826c7twe8j6299w.png)
we can use this value of
to find the corresponding value of
. From Eq(A)
![10y + 2x =520](https://img.qammunity.org/2020/formulas/mathematics/college/iud5w61gdyxfrauorqdzc9fvylvm3hj44q.png)
![10y +2(130) = 520](https://img.qammunity.org/2020/formulas/mathematics/college/wqv59qg1h79w8rg58cdkf826qgfkjxpbze.png)
![y = (520 - 2(130))/(10)](https://img.qammunity.org/2020/formulas/mathematics/college/xwmq16lrwmtuvyfzh72m4i170kwkq9723b.png)
![y = 26](https://img.qammunity.org/2020/formulas/mathematics/college/4va8usclgzw5e10x2ksa3ceo2xs98rxypb.png)
the length of the left fence will be
![y =26ft](https://img.qammunity.org/2020/formulas/mathematics/college/kwcdsqj3nl3lkr5r9cml5h8w572rvctk57.png)
- The enclosed area by the fence will be
![A = (xy)/(2)](https://img.qammunity.org/2020/formulas/mathematics/college/unzm2stz08klow26c6yeskw1hl1t0mbrr8.png)
![A = ((130)(26))/(2)](https://img.qammunity.org/2020/formulas/mathematics/college/9php92egqh5w0fb8z82vxtou1jx4kjjdml.png)
![A = 1690](https://img.qammunity.org/2020/formulas/mathematics/college/gonap30h80tuagjqnd9qrcdlfyvipiewdg.png)
Hence the maximum area that can enclosed by the fences provided the costs will be
![1690ft^2](https://img.qammunity.org/2020/formulas/mathematics/college/xztga4vuab2jhu6f2mihdgmgwvr5a6bi56.png)
- You can even check the cost of the dimensions whether they all add up to $520 or not.
Use Eq(A)
![10y + 2x =520](https://img.qammunity.org/2020/formulas/mathematics/college/iud5w61gdyxfrauorqdzc9fvylvm3hj44q.png)
![10(26) + 2(130) =520](https://img.qammunity.org/2020/formulas/mathematics/college/umdopuxbiuntld935vz22upzgkpsx4w65u.png)
and indeed it does!