Respuesta :
Answer:
[tex]500,500\text{ and }500\sqrt{2}[/tex] feet
Step-by-step explanation:
GIVEN: The boundary of a field is a right triangle with a straight stream along its hypotenuse and with fences along its other two sides.
TO FIND: Find the dimensions of the field with maximum area that can be enclosed with [tex]1000[/tex] feet of fencing.
SOLUTION:
Let the other two sides of triangle be [tex]x[/tex] and [tex]y[/tex]
such that, Â [tex]x+y=1000 \implies x=1000-y[/tex]
Area of triangle [tex]A=\frac{1}{2}\times base\times height=\frac{1}{2}\times x\times y[/tex]
Area of triangle[tex]A=\frac{1}{2}\times(1000-y)\times y=\frac{(1000y-y^2)}{2}[/tex]
to maximize area, put [tex]\frac{d\ A}{d\ y}=0[/tex]
[tex]\implies 500-y=0[/tex]
[tex]\implies y=500\text{ feet}[/tex]
other side [tex]x=1000-y=500\text{ feet}[/tex]
[tex]\text{hypotenuse}^2=x^2+y^2[/tex]
          [tex]=500^2+500^2\implies \text{hypotenuse}=500\sqrt{2}[/tex]
dimension of triangle are [tex]500,500\text{ and }500\sqrt{2}[/tex] feet
The dimensions that maximize the area of the field are 500 feet by 500 feet and the maximum area is 125000 square feet
Assume the base and the height of the right triangle field are b and h, then the following must be true:
[tex]b + h = 1000[/tex]
Make b the subject of formula
[tex]b = 1000 - h[/tex]
The area of a right-triangle is:
[tex]A = \frac 12bh[/tex]
Substitute 1000 - h for b
[tex]A = \frac 12(1000 - h)h[/tex]
Expand
[tex]A = \frac 12(1000h - h^2)[/tex]
Further, expand
[tex]A = 500h - 0.5h^2[/tex]
Differentiate
[tex]A' = 500 - h[/tex]
Set to 0
[tex]500 - h = 0[/tex]
Solve for h
[tex]h = 500[/tex]
Recall that:
[tex]b = 1000 - h[/tex]
So, we have:
[tex]b = 1000 - 500[/tex]
[tex]b = 500[/tex]
Also, recall that:
[tex]A = \frac 12bh[/tex]
So, we have:
[tex]A = \frac 12 \times 500 \times 500[/tex]
[tex]A = 125000[/tex]
Hence, the dimensions that maximize the area of the field are 500 feet by 500 feet and the maximum area is 125000 square feet
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