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3 years ago

12

A box of fruit has three times as many nectarines as grapefruit. Together there are 64 pieces of fruit. Write the equation that

represents this situation.

1 answer:

Basile [38]3 years ago

7 0

Answer:

4g = 64

Step-by-step explanation:

Let n = the number of nectarines

and g = the number of grapefruit

We have two conditions that must be satisfied to represent the situation:

(1) n = 3g

(2) n + g = 64

If you need one equation, we can substitute (1) into (2) and get

4g = 64

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A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

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Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

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