Formula is π * r^2
π = about 3.14 and the radius is half of the diameter so 28/2 = 14
3.14 * 14^2 = 615.44
B
Wrap me in plastic is a good one
The length and width of the rug is 20 and 11 feet respectively
The given parameters;
dimension of the room = 19 ft by 28 ft
maximum area of rug she can afford = 220 ft²
For a uniform stripe of floor around the rug, then suppose the uniform excess length of the floor to removed from each dimension = y
(28-2x)(19-2x)=220
532-94x+4x^2=220
4x^2-94x+312=0
x=39/2,4
For x=39/2, dimensions are negative.
The uniform dimension of the floor to be covered by the maximum area of rug she can afford = (28 - 4×2) and (19 -2×4 ) = 20 and 11
Thus, the dimensions of the rug should be 20 feet and 11 feet
- The area of a rectangle is length times breadth.
- Area is the total squares cm occupied by a closed figure.
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The maximum value of the objective function is 330
<h3>How to maximize the
objective function?</h3>
The given parameters are:
Max w = 5y₁ + 3y₂
Subject to
y₁ + y₂ ≤ 50
2y₁ + 3y₂ ≤ 60
y₁ , y₂ ≥ 0
Start by plotting the graph of the constraints (see attachment)
From the attached graph, we have:
(y₁ , y₂) = (90, -40)
Substitute (y₁ , y₂) = (90, -40) in w = 5y₁ + 3y₂
w = 5 * 90 - 3 * 40
Evaluate
w = 330
Hence, the maximum value of the function is 330
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