The area of the rectangle is 40% larger than the area of the triangle.What is the length ,L, of the rectangle?
1 answer:
Answer:
10 units
Step-by-step explanation:
Area of the triangle (A') = 1/2bh
A' = 1/2bh............ Equation 1
Where b = base of the triangle, h = height of the triangle
From the diagram,
Given: b = 12.5 unit, h = 4 unit
Substitute into equation 1
A' = 1/2(12.5)(4)
A' = 25 squared unit.
Also,
Area of the rectangle (A) = Lenght(L)×width(w)
A = Lw............ Equation 2
From the question,
If the area of the rectangle is 40% larger than the area of the triangle
Therefore,
A = 1.4A'
A = 1.4(25)
A = 35 squared unit.
Also given: w = 3 = 7/2
Substitute into equation 2
35 = L×7/2
7L = 35×2
7L = 70
L = 70/7
L = 10 units
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Answer:
Dimensions are 2 m by 1 meter by 1 meter,
Minimum cost is $ 18.
Step-by-step explanation:
Let w be the width ( in meters ) of the container,
Since, the length is twice of the width,
So, length of the container = 2w,
Now, if h be the height of the container,
Volume = length × width × height
2 = 2w × w × h
1 = w² × h
Since, the area of the base = l × w = 2w × w = 2w²,
Area of the lid = l × w = 2w²,
While the area of the sides = 2hw + 2hl
= 2h( w + l)
Since, Material for the base costs $1 per m². Material for the sides and lid costs $2 per m²,
So, the total cost,
Differentiating with respect to w,
Again differentiating with respect to w,
For maxima or minima,
C'(w) = 0
For w = 1, C''(w) = positive,
Hence, for width 1 m the cost is minimum,
Therefore, the minimum cost is C(1) = 6(1)²+12 = $ 18,
And, the dimension for which the cost is minimum is,
2 m by 1 meter by 1 meter.