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Naily [24]
3 years ago
12

Garrett and Carter went to the movies last weekend. Together they bought 2 drinks and 2 popcorns. They spent a total of $12.50.

Thursday they met several other friends at the same theater. They purchased 6 drinks and 5 popcorns for a total of $33.75. What was the cost of one drink and one popcorn?

Mathematics
1 answer:
Tatiana [17]3 years ago
4 0
Since Garrett and Carter bought popcorn and drinks, we have to figure out how much money is 1 drink and 1 popcorn. I figured the best way is to figure out half of $12.50 because 2 drinks and 2 popcorn cost that amount of money. You divide $12.50 by 2 to figure out half of that amount. The quotient equals to $6.25.

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katen-ka-za [31]

Answer:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

Step-by-step explanation:

Let the dimensions of the box be x, y and z

The rectangular box has a square base.

Therefore, Volume of the boxV=x^2z

Volume of the box=12 ft^3\\

Therefore, x^2z=12\\z=\frac{12}{x^2}

The material for the base costs \$0.17/ft^2, the material for the sides costs \$0.10/ft^2, and the material for the top costs \$0.13/ft^2.

Area of the base =x^2

Cost of the Base =\$0.17x^2

Area of the sides =4xz

Cost of the sides==\$0.10(4xz)

Area of the Top =x^2

Cost of the Base =\$0.13x^2

Total Cost, C(x,z) =0.17x^2+0.13x^2+0.10(4xz)

Substituting z=\frac{12}{x^2}

C(x) =0.17x^2+0.13x^2+0.10(4x)(\frac{12}{x^2})\\C(x)=0.3x^2+\frac{4.8}{x} \\C(x)=\dfrac{0.3x^3+4.8}{x}

To minimize C(x), we solve for the derivative and obtain its critical point

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Recall: z=\frac{12}{x^2}=\frac{12}{2^2}=3\\

Therefore, the dimensions that minimizes the cost of the box are:

(a)Length =2 feet

(b)Width =2 feet

(c)Height=3 feet

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