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andrew-mc [135]
3 years ago
13

Over what interval will the immediate value theorem apply

Mathematics
1 answer:
koban [17]3 years ago
6 0

Answer:

Any [a,b] that does NOT include the x-value 3 in it.

Either an [a,b] entirely to the left of 3, or

an  [a,b] entirely to the right of 3

Step-by-step explanation:

The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.

Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.

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The base of an aquarium with given volume V is made of slate and the sides are made of glass. If the slate costs seven times as
lubasha [3.4K]

Answer:

l = \sqrt[3]{\frac{2V}{7}}     b = \sqrt[3]{\frac{2V}{7}}       h = \sqrt[3]{\frac{49V}{4}}

Step-by-step explanation:

Represent the volume of the box with V and the dimensions with l, b and h.

The volume (V) is:

V = l * b * h

Make h the subject of the formula

h = \frac{V}{lb}

The surface area (S) of the aquarium is:

S = lb + 2(lh + bh)

Where lb represents the area of the base (i.e. slate):

The cost (C) of the surface area is:

C = 7 * lb + 1 * 2(lh + bh)

C = 7lb + 2(lh + bh)

C = 7lb + 2h(l + b)

Substitute \frac{V}{lb} for h in the above equation

C = 7lb + 2*\frac{V}{lb}(l + b)

C = 7lb + \frac{2V}{lb}(l + b)

C = 7lb + \frac{2V}{b} + \frac{2V}{l}

Differentiate with respect to l and with respect to b

C_l=7b - \frac{2V}{l^2} =0

C_b=7l - \frac{2V}{b^2} =0

To solve for b and l, we equate both equations and set l to b (to minimize the cost)

7b - \frac{2V}{l^2}=7l - \frac{2V}{b^2}

7l - \frac{2V}{l^2}=7b - \frac{2V}{b^2}

By comparison:

l =b

C_l=7b - \frac{2V}{l^2} =0 becomes

7l - \frac{2V}{l^2}=0

7l = \frac{2V}{l^2}

Cross Multiply

7l^3 = 2V

Solve for l

l^3 = \frac{2V}{7}

l = \sqrt[3]{\frac{2V}{7}}

Recall that: l =b

b = \sqrt[3]{\frac{2V}{7}}

Also recall that:

h = \frac{V}{lb}

h = \frac{V}{\sqrt[3]{\frac{2V}{7}}*\sqrt[3]{\frac{2V}{7}}}

h = \frac{V}{\sqrt[3]{\frac{4V^2}{49}}}

Apply law of indices

h = \sqrt[3]{\frac{49V^3}{4V^2}}

h = \sqrt[3]{\frac{49V}{4}}

The dimension that minimizes the cost of material of the aquarium is:

l = \sqrt[3]{\frac{2V}{7}}     b = \sqrt[3]{\frac{2V}{7}}       h = \sqrt[3]{\frac{49V}{4}}

8 0
3 years ago
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