Answer:
The length of 20 feet and width of 40 feet will result in the least amount of fencing.
Explanation:
Please find the attachment.
Let w represent width and l represent length of the rectangle.
We have been given that a rectangular area against a wall is to be fenced off on the other three sides to enclose 800 square feet.
We know that area of rectangle is width times length that is:
![A=w\cdot l](https://img.qammunity.org/2021/formulas/mathematics/high-school/kfg1fcaieksl4yg7zjgtxpt8gp4n6lj4wd.png)
This is our constraint equation.
We can see from the attachment that the fencing would be for 3 sides that is:
![\text{Perimeter}=l+l+w](https://img.qammunity.org/2021/formulas/mathematics/high-school/bepugy1ty8du4mv032rm4fxazrz3pua1zr.png)
This is our objective equation.
From constraint equation, we will get:
![w\cdot l=800](https://img.qammunity.org/2021/formulas/mathematics/high-school/2c3ce4ctklyyzjslj4xqrz040uacc9loxt.png)
![w=(800)/(l)](https://img.qammunity.org/2021/formulas/mathematics/high-school/f8vg9cse2zktnee61w4x3wouvs141tbyoz.png)
Substitute this value in objective equation:
![P=2l+(800)/(l)](https://img.qammunity.org/2021/formulas/mathematics/high-school/e9c9zco4kzq5w7elh6y37g82seeufgpagb.png)
![P=2l+800l^(-1)](https://img.qammunity.org/2021/formulas/mathematics/high-school/hn9tckpsf8yry4ypmudpjttwm5f5vp5gnf.png)
Let us find the derivative of objective equation.
![P'=2-800l^(-2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/5ud52dnfifxpchv7d3wcll07juvqe716x1.png)
Now, we will set the derivative equal to 0 to solve for length:
![2-800l^(-2)=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/2yx71oiu2o87pqtp9aquhtzkwuyvu26gkp.png)
![2-(800)/(l^2)=0](https://img.qammunity.org/2021/formulas/mathematics/high-school/vzs4yqc7vc6vxbfsad4zyadgari32xwhtv.png)
![-(800)/(l^2)=-2](https://img.qammunity.org/2021/formulas/mathematics/high-school/vztyaegchmb58ee6jvq50xlvmwap2tsbsv.png)
Cross multiply:
![-2l^2=-800](https://img.qammunity.org/2021/formulas/mathematics/high-school/3dpdsf4s8275mkyuk0tlw3vx3qsq81ogty.png)
![(-2l^2)/(-2)=(-800)/(-2)](https://img.qammunity.org/2021/formulas/mathematics/high-school/u7ikgtcfvgo03hpzqqy2364funp5dq1scu.png)
![l^2=400](https://img.qammunity.org/2021/formulas/mathematics/high-school/2m0vvdqbl00khl3kaa9emvth71btqy7bjb.png)
Take positive square root:
![l=√(400)](https://img.qammunity.org/2021/formulas/mathematics/high-school/bumd15pvj48ia078f6fo9ltfm13oshx6jb.png)
![l=20](https://img.qammunity.org/2021/formulas/mathematics/high-school/soxojrk070a1qxzjp2gt3f1w20z8vg1j7y.png)
Upon substituting
in
, we will get:
![w=(800)/(20)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ih4j22a092370yrsm252rtm1w54xz2gs25.png)
![w=40](https://img.qammunity.org/2021/formulas/mathematics/high-school/n3caamhydegq4ih5yezj6720e95sif0eb3.png)
Therefore, the length of 20 feet and width of 40 feet will result in the least amount of fencing.