To find a minimum or the maximum of a quadratic equation, we take the first derivative the equation and set it to zero (the first derivative is the slope of the equation; whenever the slope is a 0 we are at a minimum or maximum). In the case we take the first derivative of our A equation with respect to L:
A = 18 L - L²2
dA/dL = 18 - 2 L = 0
18 - 2 L = 0
18 = 2 L
L = 9 m
Now we solve for W:
W = 18 - L
W = 18 - 9
W = 9 m
Verify that the perimeter is correct:
2 W + 2 L = 36
2 (9) + 2 (9) = 36
18 + 18 = 36
36 = 36
Graph our equation A = 18 L - L².The L cannot be less than or equal to 0 since a rectangle cannot have a negative length. Also note that L cannot be greater than or equal to 18 since if it was then:
W = 18 - L = 18 - 18 = 0
which is an invalid width for a rectangle. Notice that the maximum value for A = 81 m² which is at point (9, 81). Notice that the slope of our equation dA/dL = 0 when L = 9 (see dotted red line), which means that L = 9 corresponds to either a minimum or maximum value for A and in our case L = 9 corresponds to a maximum value for A:
A = L W
A = (9 m) (9 m)
A = 81 m²
For a rectangle of width W and length L, area
A = WL [1]
Perimeter = 2W+2L = 36 [2]
Find W and L for the a maximum A.
Rearrange equation 2 to solve for W:
2W = 36 –2L
W = 18-L
Substitute 18-L for W in equation 1:
A = (18-L)L = 18L -L²
We see that A is a quadratic function of L that has a maximum but no negative minimum.
Derivative dA/dL = 18 -2L
Maximum of A occurs when dA/dL= 0:
18 -2L = 0
L = 9
Maximum area when L = 9 m and W = 9m, and area = 81 m².
Answers & Comments
Verified answer
Let W = width
Let L = length
Let A = area
So we know that the perimeter is:
2 W + 2 L = 36
W + L = 18
W = 18 - L
The area is:
A = W L
A = (18 - L) L
A = 18 L - L²
To find a minimum or the maximum of a quadratic equation, we take the first derivative the equation and set it to zero (the first derivative is the slope of the equation; whenever the slope is a 0 we are at a minimum or maximum). In the case we take the first derivative of our A equation with respect to L:
A = 18 L - L²2
dA/dL = 18 - 2 L = 0
18 - 2 L = 0
18 = 2 L
L = 9 m
Now we solve for W:
W = 18 - L
W = 18 - 9
W = 9 m
Verify that the perimeter is correct:
2 W + 2 L = 36
2 (9) + 2 (9) = 36
18 + 18 = 36
36 = 36
Graph our equation A = 18 L - L².The L cannot be less than or equal to 0 since a rectangle cannot have a negative length. Also note that L cannot be greater than or equal to 18 since if it was then:
W = 18 - L = 18 - 18 = 0
which is an invalid width for a rectangle. Notice that the maximum value for A = 81 m² which is at point (9, 81). Notice that the slope of our equation dA/dL = 0 when L = 9 (see dotted red line), which means that L = 9 corresponds to either a minimum or maximum value for A and in our case L = 9 corresponds to a maximum value for A:
A = L W
A = (9 m) (9 m)
A = 81 m²
For a rectangle of width W and length L, area
A = WL [1]
Perimeter = 2W+2L = 36 [2]
Find W and L for the a maximum A.
Rearrange equation 2 to solve for W:
2W = 36 –2L
W = 18-L
Substitute 18-L for W in equation 1:
A = (18-L)L = 18L -L²
We see that A is a quadratic function of L that has a maximum but no negative minimum.
Derivative dA/dL = 18 -2L
Maximum of A occurs when dA/dL= 0:
18 -2L = 0
L = 9
Maximum area when L = 9 m and W = 9m, and area = 81 m².
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Answer:
81m²
Step-by-step explanation:
Given:
Let x and y be the lengths
Now, we do some algebraic manipulations
Then , we plug to the area of rectangle
Area = LW
Get the first derivative equated to 0. to get the minimum/maximum
Hence , the other dimension is 9.
Let's find for the other dimension
hence , the dimension is also 9.
Now , we will get the area
Therefore , the maximum area is 81 m²