Simplify $(4x+3y)^2 + (3x-4y)^2. Answer ... It is given that $x^2 + y^2 = 1$ . Using your answer in part (i), explain why the maximum value of $(4x+3y)^2$ is 25.

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Worked solution

4(d)(i)

1. Expand each squared linear expression.

$(4x+3y)^2+(3x-4y)^2$
$=16x^2+24xy+9y^2+9x^2-24xy+16y^2$

2. Collect like terms and simplify.

$16x^2+9x^2+24xy-24xy+9y^2+16y^2$
$=25x^2+25y^2$

Answer

$25x^2+25y^2$

4(d)(ii)

1. Substitute $x^2+y^2=1$ into the result from part (i).

From part (i),
$(4x+3y)^2+(3x-4y)^2=25x^2+25y^2=25(x^2+y^2).$
Given $x^2+y^2=1$ ,
$(4x+3y)^2+(3x-4y)^2=25.$

2. Use non-negativity of a square to find the upper bound.

Since $(3x-4y)^2\geq 0$ ,
$(4x+3y)^2=25-(3x-4y)^2\leq 25.$

3. Show that the upper bound can actually be reached.

Equality occurs when
$(3x-4y)^2=0,$
so $3x-4y=0$ . This is possible together with $x^2+y^2=1$ , for example
$x=\frac45,\quad y=\frac35.$
Then
$(4x+3y)^2=\left(4\cdot\frac45+3\cdot\frac35\right)^2=5^2=25.$

Answer

Since $(4x+3y)^2+(3x-4y)^2=25$ and $(3x-4y)^2\geq 0$ , we have $(4x+3y)^2\leq 25$ . Equality is possible when $3x-4y=0$ , for example $x=\frac45$ , $y=\frac35$ . Hence the maximum value of $(4x+3y)^2$ is $25$ .

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Simplify $(4x+3y)^2 + (3x-4y)^2. Answer ... It is given that x2+y2=1x^2 + y^2 = 1x2+y2=1. Using your answer in part (i), explain why the maximum value of (4x+3y)2(4x+3y)^2(4x+3y)2 is 25.

Worked solution

4(d)(i)

4(d)(ii)

More questions on this topic

Simplify $(4x+3y)^2 + (3x-4y)^2. Answer ... It is given that $x^2 + y^2 = 1$ . Using your answer in part (i), explain why the maximum value of $(4x+3y)^2$ is 25.