Respuesta :
Answer:
Dimensions are 350/9 ft and 17.5 ft
Step-by-step explanation:
We are given the cost per ft of all the 4 sides. Let the horizontal be x and the vertical be y.
Now, we will set up the constraint and equation that we are being asked to maximize.
Thus;
700 = 10y + 10y + 7x + 2x
700 = 20y + 9x
Maiking y the subject, we have;
y = (700 - 9x)/20
y = 35 - 9x/20
Now,area of a rectangle is: A = xy
Thus, A = x(35 - 9x/20))
A = 35x - 9x²/20
We can get the critical points by finding the derivatives and Equating to zero
Thus;
dA/dx = 35 - 0.9x
At dA/dx = 0,we have; x = 350/9
At d²A/dx², we have;
d²A/dx² = -0.9
This is negative, thus we will disregard and use the one gotten from the first derivative.
Thus, we will use x = 350/9 ft
Plugging this into the equation y = 35 - 9x/20, we have;
y = 35 - ((9 × 350/9)/20)
y = 17.5 ft
The dimensions of the field that will maximize the enclosed area are 350/9 ft and 17.5 ft and this can be determined by forming the linear equation.
Given :
- We are going to fence in a rectangular field.
- The cost of the vertical sides is $10/ft, the cost of the bottom is $2/ft and the cost of the top is $7/ft.
Let 'x' be vertical, and 'y' be horizontal. So, the linear equation becomes:
700 = 10y + 10y + 7x + 2x
Simplify the above expression.
700 = 20y + 9x
Now, solve the above equation for 'y'.
[tex]\rm y = \dfrac{700-9x}{20}[/tex] --- (1)
Now, the formula of the area of the rectangle is:
A = xy
Now, substitute the value of 'y' in the above formula.
[tex]\rm A = x \times \dfrac{700-9x}{20}[/tex]
[tex]\rm A = 35x -\dfrac{9x^2}{20}[/tex]
Now, differentiate the above equation with respect to 'x' and then equate to 0.
[tex]\rm \dfrac{dA}{dx}=35-0.9x[/tex]
Now, equate the above equation to zero.
35 - 0.9x = 0
x = 350/9
Now, substitute the value of 'x' in equation (1).
[tex]\rm y = \dfrac{700-9\times \dfrac{350}{9}}{20}[/tex]
y = 17.5 ft
For more information, refer to the link given below:
https://brainly.com/question/1631786
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