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3 years ago

7

Which of the following is equivalent to 0 equals 3 x squared minus 12 x minus 15 when completing the square?

1 answer:

skad [1K]3 years ago

8 0

Answer:

x = 5, -1

Step-by-step explanation:

0 = 3x^2 - 12x - 15 or

3x^2 - 12x - 15 = 0

By solving for x, use the Quadratic Formula.

… x = 5 and/or -1

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2 years ago

X2 + 5x - 14 Select one of the factors of the quadratic expression.

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3 years ago

A rectangular box with a square base is to be constructed from material that costs $9 per ft2 for the bottom, $5 per ft2 for the

kramer

Answer:

$18.73ft^3$

Step-by-step explanation:

Let

Side of square base=x

Height of rectangular box=y

Area of square base=Area of top= $(side)^2=x^2$

Area of one side face= $l\times b=xy$

Cost of bottom=$9 per square ft

Cost of top=$5 square ft

Cost of sides=$4 per square ft

Total cost=$204

Volume of rectangular box= $V=lbh=x^2y$

Total cost= $9(x^2)+5x^2+4(4xy)=14x^2+16xy$

$204=14x^2+16xy$

$204-14x^2=16xy$

$y=\frac{204-14x^2}{16x}=\frac{102-7x^2}{8x}$

Substitute the values of y

$V(x)=x^2\times \frac{102-7x^2}{8x}=\frac{1}{8}(102x-7x^3)$

Differentiate w.r.t x

$V'(x)=\frac{1}{8}(102-21x^2)=0$

$V'(x)=0$

$\frac{1}{8}(102-21x^2)=0$

$102-21x^2=0$

$102=21x^2$

$x^2=\frac{102}{21}=4.85$

$x=\sqrt{4.85}=2.2$

It takes positive because side length cannot be negative.

Again differentiate w.r. t x

$V''(x)=\frac{1}{8}(-42x)$

Substitute the value

$V''(2.2)=-\frac{42}{8}(2.2)=-11.55$

Hence, the volume of box is maximum at x=2.2 ft

Substitute the value of x

$y=\frac{102-7(2.2)^2}{8(2.2)}=3.87 ft$

Greatest volume of box= $x^2y=(2.2)^2\times 3.87=18.73 ft^3$

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4 years ago